The value of int e^{tan ^{-1} x} cdot frac{1+ | vedclass

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(D) Let $I = \int e^{	an ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$. We can rewrite the integrand as: $I = \int e^{	an ^{-1} x} \left( \frac{1+x^2}{1+

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

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(D) Let $I = \int e^{	an ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$. We can rewrite the integrand as: $I = \int e^{	an ^{-1} x} \left( \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} ight) dx$ $I = \int e^{	an ^{-1} x} \left( 1 + \frac{x}{1+x^2} ight) dx$ $I = \int e^{	an ^{-1} x} dx + \int e^{	an ^{-1} x} \cdot \frac{x}{1+x^2} dx$. Using integration by parts on the first integral $\int e^{	an ^{-1} x} \cdot 1 dx$: Let $u = e^{	an ^{-1} x}$ and $dv = dx$. Then $du = e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$. Applying the formula $\int u dv = uv - \int v du$: $\int e^{	an ^{-1} x} dx = x e^{	an ^{-1} x} - \int x \cdot e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} dx$. Substituting this back into the expression for $I$: $I = \left( x e^{	an ^{-1} x} - \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx ight) + \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx + c$. $I = x e^{	an ^{-1} x} + c$.

The value of $\int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$ is

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