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

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(A) Let $I = \int {{e^{{{	an }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} ight)dx$. Substitute ${	an ^{ - 1}}x = t$,which implies $x

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$x{e^{{{	an }^{ - 1}}x}} + c$

Vedclass · Answer

(A) Let $I = \int {{e^{{{	an }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} ight)dx$. Substitute ${	an ^{ - 1}}x = t$,which implies $x = 	an t$ and $dx = \sec^2 t \, dt$. Also,$\frac{dx}{1+x^2} = dt$. Substituting these into the integral: $I = \int e^t (1 + 	an t + 	an^2 t) dt$ $I = \int e^t (1 + 	an^2 t + 	an t) dt$ Since $1 + 	an^2 t = \sec^2 t$,we have: $I = \int e^t (\sec^2 t + 	an t) dt$. Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = 	an t$ and $f'(t) = \sec^2 t$: $I = e^t 	an t + C$. Substituting back $t = 	an^{-1} x$: $I = x e^{	an^{-1} x} + C$.

$\int {{e^{{{\tan }^{ - 1}}x}}} \left( {\frac{{1 + x + {x^2}}}{{1 + {x^2}}}} \right)dx$ is equal to

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