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

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(A) Let $I = \int e^{	an ^{-1} x} \left(1 + \frac{x}{1+x^2}ight) dx$.We can split the integral as $I = \int e^{	an ^{-1} x} dx + \int \frac{x e^{\

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

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(A) Let $I = \int e^{	an ^{-1} x} \left(1 + \frac{x}{1+x^2}ight) dx$.We can split the integral as $I = \int e^{	an ^{-1} x} dx + \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx$.Now,apply integration by parts to the first integral $\int e^{	an ^{-1} x} dx$ by taking $u = e^{	an ^{-1} x}$ and $dv = dx$.Then $du = e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$.Using the formula $\int u dv = uv - \int v du$,we get:$\int e^{	an ^{-1} x} dx = x e^{	an ^{-1} x} - \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx$.Substituting this back into the expression for $I$:$I = \left(x e^{	an ^{-1} x} - \int \frac{x e^{	an ^{-1} x}}{1+x^2} dxight) + \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx + c$.The integral terms cancel out,leaving $I = x e^{	an ^{-1} x} + c$.

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

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