Evaluate the integral: int {frac{{{e^{{{tan } | vedclass

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Question

(C) Given integral is $I = \int {\frac{{{e^{{{	an }^{ - 1}}x}}}}{{(1 + {x^2})}}\,\,\left[ {{{\left( {{{\sec }^{ - 1}}\,\sqrt {1 + {x^2}} } ight)}^2

Vedclass · Accepted Answer

${e^{{{	an }^{ - 1}}x}}\,.\,{\left( {{{	an }^{ - 1}}x} ight)^2}\,\,+ C$

Vedclass · Answer

(C) Given integral is $I = \int {\frac{{{e^{{{	an }^{ - 1}}x}}}}{{(1 + {x^2})}}\,\,\left[ {{{\left( {{{\sec }^{ - 1}}\,\sqrt {1 + {x^2}} } ight)}^2}\,\, + \,\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight)} ight]} \,\,\,dx$. First,simplify the trigonometric expressions: For $x > 0$,let $	an^{-1}x = 	heta$,then $x = 	an	heta$. $\sec^{-1}\sqrt{1+x^2} = \sec^{-1}(\sec	heta) = 	heta = 	an^{-1}x$. Also,$\cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = \cos^{-1}(\cos 2	heta) = 2	heta = 2	an^{-1}x$. Substituting these into the integral: $I = \int \frac{e^{	an^{-1}x}}{1+x^2} [(	an^{-1}x)^2 + 2	an^{-1}x] dx$. Let $t = 	an^{-1}x$,then $dt = \frac{1}{1+x^2} dx$. $I = \int e^t (t^2 + 2t) dt$. Using the formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$,where $f(t) = t^2$ and $f'(t) = 2t$: $I = e^t \cdot t^2 + C = e^{	an^{-1}x} (	an^{-1}x)^2 + C$.

Evaluate the integral: $\int {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{(1 + {x^2})}}\,\,\left[ {{{\left( {{{\sec }^{ - 1}}\,\sqrt {1 + {x^2}} } \right)}^2}\,\, + \,\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right]} \,\,\,dx$ for $x > 0$.

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