int e^{tan^{-1} x} left( frac{1+x+x^2}{1+x^2} | vedclass

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(C) Let $u = 	an^{-1} x$. Then $x = 	an u$ and $dx = \sec^2 u \, du = (1+x^2) \, du$.Substituting these into the integral:$I = \int e^u \left( \frac

Vedclass · Accepted Answer

$x \cdot e^{	an^{-1} x}$

Vedclass · Answer

(C) Let $u = 	an^{-1} x$. Then $x = 	an u$ and $dx = \sec^2 u \, du = (1+x^2) \, du$.Substituting these into the integral:$I = \int e^u \left( \frac{1+x+x^2}{1+x^2} ight) (1+x^2) \, du = \int e^u (1+x+x^2) \, du$.Since $x = 	an u$,we have $1+x^2 = \sec^2 u$.So,$I = \int e^u (1 + 	an u + 	an^2 u) \, du = \int e^u (\sec^2 u + 	an u) \, du$.We know that $\frac{d}{du}(	an u) = \sec^2 u$.Using the standard integral formula $\int e^u (f(u) + f'(u)) \, du = e^u f(u) + C$,where $f(u) = 	an u$ and $f'(u) = \sec^2 u$,we get:$I = e^u 	an u + C = e^{	an^{-1} x} \cdot x + C$.

$\int e^{\tan^{-1} x} \left( \frac{1+x+x^2}{1+x^2} \right) dx = \rule{1cm}{0.15mm} + C$

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