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

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

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

Vedclass · Answer

(B) Let $I = \int e^{	an ^{-1} x} \left(1 + \frac{x}{1 + x^2} ight) dx$. Substitute $t = 	an ^{-1} x$,which implies $x = 	an t$ and $dx = (1 + x^2) dt = \sec^2 t dt$. Then,$I = \int e^t \left(1 + \frac{	an t}{1 + 	an^2 t} ight) \sec^2 t dt$. Since $1 + 	an^2 t = \sec^2 t$,the expression becomes $I = \int e^t \left(1 + \frac{	an t}{\sec^2 t} ight) \sec^2 t dt$. $I = \int e^t (\sec^2 t + 	an t) dt$. Using the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = 	an t$ and $f'(t) = \sec^2 t$. Thus,$I = e^t 	an t + c$. Substituting back $t = 	an ^{-1} x$,we get $I = x e^{	an ^{-1} x} + c$.

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

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