12th NCERT INTEGRALS Extra questions
Do or do not
There is no try

Question (1)

$I = \int {\frac{1}{{{x^4} + 1}}dx}$

Solution

$I = \int {\frac{1}{{{x^4} + 1}}dx}$ $= \frac{1}{2}\int {\frac{{\left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right)}}{{{x^4} + 1}}} dx$ $= \frac{1}{2}\int {\frac{{{x^2} + 1}}{{{x^4} + 1}}dx - \frac{1}{2}\int {\frac{{{x^2} - 1}}{{{x^4} + 1}}dx} }$ $Divide\;by\;{x^2}$ $= \frac{1}{2}\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}dx - \frac{1}{2}\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}dx} }$ $= \frac{1}{2}\int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}dx - \frac{1}{2}\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 2}}dx} }$ $Let\;x - \frac{1}{x} = {\mathop{\rm t}\nolimits} \;and\;x + \frac{1}{x} = u$ $\therefore \left( {1 + \frac{1}{{{x^2}}}} \right) = dt\;and\;\left( {1 - \frac{1}{{{x^2}}}} \right) = du$ $I = \;\frac{1}{2}\int {\frac{{dt}}{{{t^2} + 2}} - \frac{1}{2}\int {\frac{{du}}{{{u^2} - 2}}} }$ $I = \frac{1}{2}\int {\frac{{dt}}{{{{\left( t \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}} - \frac{1}{2}\int {\frac{{du}}{{{{\left( u \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}} }$ $I = \frac{1}{2}\left[ {\frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} \right)} \right] - \frac{1}{2}\left[ {\frac{1}{{2\sqrt 2 }}\log \left| {\frac{{u - \sqrt 2 }}{{u + \sqrt 2 }}} \right|} \right] + c$ $I = \frac{1}{{2\sqrt 2 }}{\tan ^{ - 1}}\left[ {\frac{1}{{\sqrt 2 }}\left( {x - \frac{1}{x}} \right)} \right] - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{x + \frac{1}{x} - \sqrt 2 }}{{x + \frac{1}{x} + \sqrt 2 }}} \right| + c$ $I = \frac{1}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{{x^2} + 1 - \sqrt 2 x}}{{{x^2} + 1 + \sqrt 2 x}}} \right| + c$