12th NCERT INTEGRALS Exercise 7.6 Questions 24
Do or do not
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Integrate the rational functions in Exercise 1 to 21

Question (1)

xsinx

Solution

\[I = \int {x\sin xdx} \]
In question x is algebraic while sinx is trigonometric
so x = u and sinx = v.dx
Use following formula
\[\int {uvdx = u\int {vdx - \int {\left( {\frac{d}{{dx}}u\int {vdx} } \right)} } } dx\]
\[I = \int {x\sin xdx} \] \[I = x\int {\sin xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin xdx} } \right)} } dx\] \[I = x\left( { - \cos x} \right) - \int {1\left( { - \cos x} \right)dx} \] \[I = - x\cos x + \int {\cos xdx} \] \[I = - x\cos x + sinx + c\]

Question (2)

xsin3x

Solution

\[I = \int {x\sin 3xdx} \]
In question x is algebraic while sin3x is trigonometric
so x = u and sin3x = v.dx
Use following formula
\[\int {uvdx = u\int {vdx - \int {\left( {\frac{d}{{dx}}u\int {vdx} } \right)} } } dx\]
\[I = x\int {\sin 3xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin 3xdx} } \right)} } dx\]
\[\int {\sin ax = \frac{{ - \cos ax}}{a}} \]
\[I = x\left( {\frac{{ - \cos 3x}}{3}} \right) - \int {1\left( {\frac{{ - \cos 3x}}{3}} \right)dx} \] \[I = - x\frac{{\cos 3x}}{3} + \frac{1}{3}\int {\cos 3xdx} \] \[I = - \frac{x}{3}\cos x + \frac{1}{9}sin3x + c\]

Question (3)

x2ex

Solution

\[I = \int {{x^2}{e^x}dx} \]
In question x2 is algebraic while ex is exponential
so x2 = u and ex = v.dx
Use following formula
\[\int {uvdx = u\int {vdx - \int {\left( {\frac{d}{{dx}}u\int {vdx} } \right)} } } dx\]
\[I = {x^2}\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}{x^2}\int {{e^x}dx} } \right)} } dx\] \[I = {x^2}{e^x} - \int {2x\left( {{e^x}} \right)dx} \]
Again x is algebraic while ex is exponential
so x = u and ex = v.dx
\[I = {x^2}{e^x} - 2\left[ {x\int {{e^x}dx - \int {\frac{d}{{dx}}x\int {{e^x}} dx} } } \right]\] \[I = {x^2}{e^x} - 2x{e^x} + 2{e^x} + c\] \[I = {e^x}\left( {{x^2} - 2x + 2} \right) + c\]

Question (4)

xlogx

Solution

\[I = \int {x\log xdx} \]
Here x is algebraic and logx is logarithmic function
According to order u=logx and vdx= xdx. Use formula after rewriting equation
\[\int {uvdx = u\int {vdx - \int {\left( {\frac{d}{{dx}}u\int {vdx} } \right)} } } dx\]
\[I = \int {\log x \cdot xdx} \] \[I = \log x\int {xdx - \int {\left( {\frac{d}{{dx}}\log x\int {xdx} } \right)} } dx\] \[I = \log x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{\require{cancel}\cancel{x}} \cdot \frac{{{\cancel{x^2}^x}}}{2}dx} \] \[I = \frac{{{x^2}\log x}}{2} - \frac{1}{2}\int {xdx} \] \[I = \frac{{{x^2}\log x}}{2} - \frac{{{x^2}}}{4} + c\]

Question (5)

xlog2x

Solution

\[I = \int {x\log 2xdx} \]
Here x is algebraic and logx is logarithmic function
According to order u=log2x and vdx= xdx. Use formula after rewriting equation
\[\int {uvdx = u\int {vdx - \int {\left( {\frac{d}{{dx}}u\int {vdx} } \right)} } } dx\]
\[I = \int {\log 2x \cdot xdx} \] \[I = \log 2x\int {xdx - \int {\left( {\frac{d}{{dx}}\log 2x\int {xdx} } \right)} } dx\] \[I = \log 2x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{{\cancel{2x}}} \times \cancel{2} \cdot \frac{{{\cancel{x^2}x}}}{2}dx} \] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{1}{2}\int {xdx} \] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{1}{2}\frac{{{x^2}}}{2} + c\] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{{{x^2}}}{4} + c\]

Question (6)

x2logx dx

Solution

\[I = \int {{x^2}\log xdx} \]
Here x2 is algebraic and logx is logarithmic.
According to order u=logx and v=x2.
\[I = \int {\log x \cdot {x^2}dx} \] \[I = \log x\int {{x^2}dx} - \int {\left( {\frac{d}{{dx}}\log x\int {{x^2}dx} } \right)} dx\] \[I = \log x\frac{{{x^3}}}{3} - \int {\frac{1}{x}} \cdot \frac{{{x^3}}}{3}dx\] \[I = \frac{{{x^3}\log x}}{3} - \frac{1}{3}\int {{x^2}} dx\] \[I = \frac{{{x^3}\log x}}{3} - \frac{1}{3}\frac{{{x^3}}}{3} + c\] \[I = \frac{{{x^3}\log x}}{3} - \frac{{{x^3}}}{9} + c\]

Question (7)

xsin-1x

Solution

\[I = \int {x{{\sin }^{ - 1}}xdx} \]
Here x algebraic and sin-1 is inverse.
According to order u=sin-1 and v=x.
\[I = \int {{{\sin }^{ - 1}}x \cdot xdx} \] \[I = {\sin ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \int {\left( {\frac{1}{{\sqrt {1 - {x^2}} }} \times \frac{{{x^2}}}{2}} \right)} dx\] \[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\qquad equation(1) \] \[\qquad Let{I_1} = \int {\left( {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] Add 1 and subtract 1 from Numerator
\[{I_1} = \int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[{I_1} = \int {\left( {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx + \int {\left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[{I_1} = \int {\sqrt {1 - {x^2}} dx - {{\sin }^{ - 1}}x + c} \] \[{I_1} = \frac{{{x}}}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}x - {\sin ^{ - 1}}x + c\] \[{I_1} = \frac{{{x}}}{2}\sqrt {1 - {x^2}} - \frac{1}{2}{\sin ^{ - 1}}x + c\] Sbustituting value of I1 n equation(1)
\[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\qquad equation(1) \] \[I = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\left[ {\frac{x}{2}\sqrt {1 - {x^2}} - \frac{{{{\sin }^{ - 1}}x}}{2} + c} \right]\] \[I = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{4} - \frac{{{{\sin }^{ - 1}}x}}{4} + c\] \[I = \frac{{{{\sin }^{ - 1}}x}}{2}\left( {{x^2} - \frac{1}{2}} \right) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c\] \[I = \frac{{\left( {2{x^2} - 1} \right){{\sin }^{ - 1}}x}}{4} + \frac{{x\sqrt {1 - {x^2}} }}{4} + c\]

Question (8)

xtan-1x

Solution

Here x is algebraic and tan-1x is inverse
According to order u=tan-1x and v = x
\[I = \int {x{{\tan }^{ - 1}}} xdx\] \[I = \int {{{\tan }^{ - 1}}x \cdot xdx} \] \[I = {\tan ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = {\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{{1 + {x^2}}} \times \frac{{{x^2}}}{2}dx} \] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} \] Add and subtract 1 in numerator
\[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{1 + {x^2} - 1}}{{1 + {x^2}}}dx} \] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{1 + {x^2}}}{{1 + {x^2}}}dx} + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx\] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {dx} + \frac{1}{2}{\tan ^{ - 1}}x + c\] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{x}{2} + \frac{{{{\tan }^{ - 1}}x}}{2} + c\]

Question (9)

xcos-1

Solution

\[I = \int {x{{\cos }^{ - 1}}xdx} \]
Here x algebraic and cos-1 is trigonometric.
According to order u=cos-1 and v=x.
\[I = \int {{{\cos }^{ - 1}}x \cdot xdx} \] \[I = {\cos ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\cos }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \int {\left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }} \times \frac{{{x^2}}}{2}} \right)} dx\] \[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] Add 1 and subtract 1 from Numerator
\[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx + \frac{1}{2}\int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} \] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\sqrt {1 - {x^2}} dx} + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\left[ {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x} \right] + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} - \frac{1}{4}{\sin ^{ - 1}}x + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{1}{4}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{1}{4}\left( {\frac{\pi }{2} - {{\cos }^{ - 1}}x} \right) + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{\pi }{8} - \frac{{{{\cos }^{ - 1}}x}}{4} + {c_1}\] \[c = {c_1} + \frac{\pi }{8}\] \[I = \frac{{\left( {2{x^2} - 1} \right){{\cos }^{ - 1}}x}}{4} - \frac{{x\sqrt {1 - {x^2}} }}{4} + c\]

Question (10)

(sin-1x)2

Solution

\[I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx} \]
Here sin-1x is inverse function. It has only one function so we will take 1 as other function which is algebraic. According to order
u= (sin-1x)2 and v=1
\[I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} \cdot 1dx} \] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}\int {1dx} - \int {\left( {\frac{d}{{dx}}{{\left( {{{\sin }^{ - 1}}x} \right)}^2}\int {1dx} } \right)} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x - \int {\left( {2{{\sin }^{ - 1}}x} \right)\frac{1}{{\sqrt {1 - {x^2}} }}x} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x - 2\int {\frac{{{{\sin }^{ - 1}}x \cdot x}}{{\sqrt {1 - {x^2}} }}} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x + {I_1}\qquad equation(1)\] \[let\qquad {I_1} = \int {\frac{{{{\sin }^{ - 1}}x \cdot x}}{{\sqrt {1 - {x^2}} }}} dx\] Let sin-1x = θ & x = sinθ \[\therefore \frac{1}{{\sqrt {1 - {x^2}} }}dx = d\theta \] \[{I_1} = \int {\theta \sin \theta \cdot d\theta } \]
Here θ θ = algebraic and sinθ is trigonometric. According to order u=θ and v=sinθ
\[I_1 = \theta \int {\sin \theta \cdot d\theta } - \int {\left( {\frac{d}{{dx}}\theta \int {\sin \theta \cdot d\theta } } \right)} d\theta \] \[I_1 = \theta \left( { - \cos \theta } \right) - \int {1\left( { - \cos \theta } \right)} d\theta \] \[{I_1} = - \theta \cos \theta + \sin \theta + c\] \[{I_1} = - \theta \sqrt {1 - {{\sin }^2}\theta } + \sin \theta + c\] \[{I_1} = - \theta \sqrt {1 - {{\sin }^2}\theta } + \sin \theta + c\] substitute value of θ
\[I_1 = - {\sin ^{ - 1}}x\sqrt {1 - {x^2}} + x + c\] Replacing value of I1 in equation (1)
\[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} - 2\left[ { - {{\sin }^{ - 1}}x\sqrt {1 - {x^2}} + x} \right] + c\] \[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2{\sin ^{ - 1}}x\sqrt {1 - {x^2}} - 2x + c\] \[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c\]

Question (11)

\[\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\]

Solution

\[I = \int {\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} \] Let cos-1x = θ ⇒ x = cosθ
taking derivative
\[\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx = d\theta \] \[\frac{1}{{\sqrt {1 - {x^2}} }}dx = - d\theta \] Replace in given equation
\[I = \int {\cos \theta \cdot \theta \left( { - d\theta } \right)} \] \[I = - \int {\cos \theta \cdot \theta d\theta } \]
Here θ is algebraic and cosθ is trigonometric
u= θ and v=cosθ
\[I = - \int {\theta \cdot \cos \theta d\theta } \] \[I = - \left[ {\theta \int {\cos \theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \int {\cos \theta d\theta } } \right)d\theta } } } \right]\] \[I = - \left[ {\theta \sin \theta - \int {1 \cdot \sin \theta d\theta } } \right]\] \[I = - \theta \sin \theta + \left( { - \cos \theta } \right) + c\] \[I = - \theta \sqrt {1 - {{\cos }^2}\theta } - \cos \theta + c\] substitute value of θ \[I = - {\cos ^{ - 1}}x\sqrt {1 - {x^2}} - x + c\] \[I = - \left[ {\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + x} \right] + c\]

Question (12)

xsec2

Solution

\[I = \int {x{{\sec }^2}xdx} \]
Here x is algebraic and sec2x is trigonometric function. According to to order u=x amd v=sec2
\[I = x\int {{{\sec }^2}xdx - \int {\left( {\frac{d}{{dx}}x\int {{{\sec }^2}xdx} } \right)} } dx\] \[I = x\tan x - \int {1 \cdot \tan xdx} \] \[I = x\tan x - \left( { - \log \left| {\cos x} \right|} \right) + c\] \[I = x\tan x + \log \left| {\cos x} \right| + c\]

Question (13)

tan-1x

Solution

\[I = \int {{{\tan }^{ - 1}}xdx} \]
Here tan-1 is inverse only one function. So we take 1 as other function which is algebraic. According to order u=tan-1 and v=1
\[I = {\tan ^{ - 1}}x\int {1dx - \int {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x \cdot \int {1dx} } \right)} } dx\] \[I = {\tan ^{ - 1}}x \cdot x - \int {\frac{1}{{1 + {x^2}}} \cdot } xdx\] \[I = x{\tan ^{ - 1}}x - \int {\frac{x}{{1 + {x^2}}}} dx\] I=x tan-1 - I1 equation(1)
\[Here\quad {I_1} = \int {\frac{x}{{1 + {x^2}}}dx} \] Let 1+x2 = t
2xdx= dt
xdx= dt/2
\[{I_1} = \int {\frac{{\frac{{dt}}{2}}}{t}} \] \[{I_1} = \frac{1}{2}\int {\frac{{dt}}{t}} \] \[{I_1} = \frac{1}{2}\log \left| t \right| + c\] \[{I_1} = \frac{1}{2}\left| {1 + {x^2}} \right| + c\] Replacing values in equation(1)
\[I = x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {1 + {x^2}} \right| + c\]

Question (14)

x(logx)2

Solution

\[I = \int {x{{\left( {\log x} \right)}^2}dx} \]
Here x is algebraic (logx)2 logarithmic
\[I = \int {{{\left( {\log x} \right)}^2} \cdot xdx} \] \[I = {\left( {\log x} \right)^2}\int {xdx - \int {\left( {\frac{d}{{dx}}{{\left( {\log x} \right)}^2}\int {xdx} } \right)} } dx\] \[I = {\left( {\log x} \right)^2}\frac{{{x^2}}}{2} - \int {\cancel{2}\left( {\log x} \right) \cdot \frac{1}{x} \cdot \frac{{{x^2}}}{\cancel{2}}dx} \] \[I = \frac{{{{\left( {\log x} \right)}^2}{x^2}}}{2} - \int {\log x \cdot xdx} \] \[Let\quad I = \frac{{{{\left( {\log x} \right)}^2}{x^2}}}{2} - {I_1}\qquad equation(1)\] \[Here\quad {I_1} = \int {\log x \cdot xdx} \] Agian using parts we get \[{I_1} = \log x\int {xdx - \int {\left( {\frac{d}{{dx}}\log x\int {xdx} } \right)dx} } \] \[{I_1} = \log x\frac{{{x^2}}}{2} - \int {\frac{1}{\cancel{x}} \cdot \frac{{{x^\cancel{2}}}}{2}dx} \] \[{I_1} = \frac{{{x^2}\log x}}{2} - \frac{1}{2} \cdot \frac{{{x^2}}}{2} + c\] \[{I_1} = \frac{{{x^2}\log x}}{2} - \frac{{{x^2}}}{4} + c\] Substitute value of I1 n equation (1)
\[I = \frac{{{x^2}{{\left( {\log x} \right)}^2}}}{2} - \frac{{{x^2}\log x}}{2} + \frac{{{x^2}}}{4} + c\]

Question (15)

(x2+1)logx

Solution

\[I = \int {\left( {{x^2} + 1} \right)\log xdx} \]
Here x2+1 s algebraic function and logx is algebraic function u = logx and v = x2+1.
\[I = \int {\log x \cdot \left( {{x^2} + 1} \right)dx} \] \[I = \log x\int {\left( {{x^2} + 1} \right)dx - \int {\left( {\frac{d}{{dx}}\log x\int {\left( {{x^2} + 1} \right)dx} } \right)} } dx\] \[I = \log x\left( {\frac{{{x^3}}}{3} + x} \right) - \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)} dx\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {\frac{1}{\cancel{x}} \cdot \cancel{x}\left( {{x^2} + 3} \right)} dx\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {{x^2}dx - \frac{3}{3}\int {dx} } \] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\frac{{{x^3}}}{3} - x + c\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{9}{x^3} - x + c\]

Question (16)

ex(sinx + cosx)

Solution

\[I = \int {{e^x}} \left( {\sin x + \cos x} \right)dx\]
When in integral we have ex is one of the function, and other function is in form of addition of two. Check whether other is derivative or not . if it is in that form then we can use the formula \[\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c\]
$$I = \int {{e^x}} \left( {\sin x + \cos x} \right)dx$$ Let sin x = f(x)
then f'(x) = cos x
\[I = \int {{e^x}} \left[ {\sin x + \cos x} \right]dx\] \[I = \int {{e^x}} \left[ {f\left( x \right) + {f'}\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\sin x + c\]

Question (17)

\[\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}\]

Solution

Here ex is exponential other is not form of addition of two function. we will try to represent as addition of two where other is derivative of two where other is derivative of first one
\[I = \int {\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}} dx\] \[I = \int {{e^x}} \left[ {\frac{x}{{{{\left( {1 + x} \right)}^2}}}} \right]dx\] \[I = \int {{e^x}} \left[ {\frac{{1 + x - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right]dx\] \[I = \int {{e^x}\left[ {\frac{{\left(\require{cancel}\cancel{ {1 + x}} \right)}}{{{{\left( {1 + x} \right)}^\cancel{2}}}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[I = \int {{e^x}\left[ {\frac{1}{{\left( {1 + x} \right)}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[{\rm{Let }}\;f\left( x \right) = \frac{1}{{1 + x}}\] \[\therefore \quad f'\left( x \right) = \frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}\] \[I = \int {{e^x}\left[ {\frac{1}{{1 + x}} + \frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[I = \int {e\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\frac{1}{{1 + x}} + c\] \[I = \frac{{{e^x}}}{{1 + x}} + c\]

Question (18)

\[{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}\]

Solution

\[I = \int {{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}dx} \]
use \[1 + \cos x = 2{\cos ^2}\frac{x}{2}\]
\[I = \int {{e^x}\left( {\frac{1}{{2{{\cos }^2}\frac{x}{2}}} + \frac{{\sin x}}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx\]
replace \[\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}\]
\[I = \int {{e^x}\left( {\frac{{{{\sec }^2}\frac{x}{2}}}{2} + \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx\] \[I = \int {{e^x}\left[ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right]} dx\] \[I = \int {{e^x}\left( {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right)} dx\] \[{\rm{Let}}\quad {\rm{f}}\left( x \right) = \tan \frac{x}{2}\] \[f'\left( x \right) = {\sec ^2}\frac{x}{2} \cdot \frac{1}{2}\] \[I = \int {{e^x}\left[ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right]} dx\] \[I = \int {{e^x}\left[ {f'\left( x \right) + f\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\tan \frac{x}{2} + c\]

Question (19)

\[{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\]

Solution

\[{\rm{Let f}}\left( {\rm{x}} \right) = \frac{1}{x}\] \[f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}\] \[\int {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx} \] \[I = \int {{e^x}} \left[ {f\left( x \right) + f'\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\frac{1}{x} + c\] \[I = \frac{{{e^x}}}{x} + c\]

Question (20)

\[\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}\]

Solution

\[I = \int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx} \]
\[\text{For function}\qquad \frac{{\left( {x - 3} \right)}}{{{{\left( {x - 1} \right)}^3}}}\text{we will try to break into two algebraic }\] functions such that one will be derivative of the other
\[\frac{{\left( {x - 3} \right)}}{{{{\left( {x - 1} \right)}^3}}} = \frac{{x - 1 - 2}}{{{{\left( {x - 1} \right)}^3}}}\] Now \[I = \int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx} \] \[I = \int {{e^x}\left[ {\frac{{x - 1 - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]dx} \] \[I = \int {{e^x}\left[ {\frac{{x - 1}}{{{{\left( {x - 1} \right)}^3}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx\] \[I = \int {{e^x}} \left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]dx\] \[\text{let}\quad f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}} = {\left( {x - 1} \right)^{-2}}\] \[f'\left( x \right) = - 2{\left( {x - 1} \right)^{ - 3}}\] \[f'\left( x \right) = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}\] \[\text{So}\quad I = \int {{e^x}\left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx\] \[I = \int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c\]

Question (21)

\[{e^{2x}}\sin x\]

Solution

\[I = \int {{e^{2x}}\sin x\;dx} \]
Here e2x is exponential function and sinx is trigonometric function. We can use integration by parts
u=snx and v=e2x
\[I = \int {\sin x\;{e^{2x}}\;dx} \] \[I = \sin x\int {{e^{2x}}dx - \int {\left( {\frac{d}{{dx}}\sin x\int {{e^{2x}}dx} } \right)} } dx\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \int {\cos x} \;\frac{{{e^{2x}}}}{2}dx\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\int {\cos x} \;{e^{2x}}dx\]
Again apply partial fraction for \[{I_1} = \int {\cos x} \;\frac{{{e^{2x}}}}{2}dx\]
\[{I_1} = \left[ {\cos x\int {{e^{2x}}dx - \int {\left( {\frac{d}{{dx}}cox\int {{e^{2x}}dx} } \right)dx} } } \right]\] \[{I_1} = \cos x\cdot\frac{{{e^{2x}}}}{2} - \int {\left( { - sinx} \right)} \frac{{{e^{2x}}}}{2}dx\] \[{I_1} = \frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}\int {\sin x\cdot{e^{2x}}dx} \] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {{I_1}} \right]\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {\frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}\int {\sin x\cdot{e^{2x}}dx} } \right]\] Here ∫sinx e2x dx is again initial question can be replaced by I \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {\frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}I} \right] + c\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} - \frac{1}{4}I + c\] \[I + \frac{1}{4}I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} + c\] \[\frac{5}{4}I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} + c\] \[I = \frac{\cancel{4}}{5}\left[ {\frac{1}{\cancel{4}}\left( {2\sin x - \cos x} \right){e^{2x}}} \right] + c\] \[I = \frac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + c\]

Question (22)

\[{\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\]

Solution

\[I = \int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} \] Let x = tanθ ⇒ θ = tan-1x
∴ dx = sec2θ dθ
Replacing the value we get \[I = \int {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)} \cdot {\sec ^2}\theta d\theta \]
Use identity \[\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta \]
\[I = \int {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right)} \cdot {\sec ^2}\theta d\theta \] \[I = \int {2\theta } {\sec ^2}\theta d\theta \] \[I = 2\int \theta {\sec ^2}\theta d\theta \]
Here θ is algebraic and sex2θ is trigonometric
According to order u=θ and v = sec2θ
\[I = 2\left[ {\theta \int {{{\sec }^2}\theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \int {{{\sec }^2}\theta d\theta } } \right)d\theta } } } \right]\] \[I = 2\left[ {\theta \tan \theta - \int {1 \cdot \tan \theta d\theta } } \right]\] \[I = 2\theta \tan \theta - 2\log \left| {\sec \theta } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log \left| {\sqrt {1 + {{\tan }^2}\theta } } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log \left| {\sqrt {1 + {x^2}} } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log {\left( {1 + {x^2}} \right)^{\frac{1}{2}}} + c\] \[I = 2x{\tan ^{ - 1}}x - \cancel{2} \cdot \frac{1}{\cancel{2}}\log \left( {1 +{x^2}} \right) + c\] \[I = 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + c\]

Choose the correct answer in Exercise 23 and 24

Question (23)

\[\int {{x^2}{e^{{x^3}}}} dx\qquad \text{equals}\] Options
\[\left( A \right)\quad \frac{1}{3}{e^{{x^3}}} + c\] \[\left( B \right)\quad\frac{1}{3}{e^{{x^2}}} + c\] \[\left( C \right)\quad\frac{1}{2}{e^{{x^3}}} + c\] \[\left( D \right)\quad\frac{1}{2}{e^{{x^2}}} + c\]

Solution

\[I = \int {{x^2}{e^{{x^3}}}dx} \] \[I = \int {{e^{{x^3}}}{x^2}dx} \] Let x3 = t
3x2dx = dt
x2dx = dt/3
\[I = \int {{e^t}} \frac{{dt}}{3}\] \[I = \frac{1}{3}\int {{e^t}dt} \] \[I = \frac{1}{3}{e^t} + c\] \[I = \frac{1}{3}{e^{{x^3}}} + c\] A is correct option

Question (24)

\[\int {{e^x}\sec x\left( {1 + \tan x} \right)} dx\] Options
(A) ex cosx + c
(B) ex secx + c
(C) ex sinx + c
(D) ex tanx + c

Solution

\[I = \int {{e^x}\sec x} \left( {1 + \tan x} \right)dx\] \[I = \int {{e^x}\left( {\sec x + \sec x\tan x} \right)} dx\] Let f(x) = secx f'(x) = secx tanx
\[I = \int {{e^x}} \left[ {f\left( x \right) + f'\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\sec x + c\] B is correct option
Exercise 7.5 ⇐
⇒ Exercise 7.7