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  4. The value of the integral ∫ ex ((1-x/1+x2))2 dx is

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Q. The value of the integral $ \int e^{x} \left(\frac{1-x}{1+x^{2}}\right)^{^2}$ dx is

VITEEEVITEEE 2007

Solution:

Let I =$ \int e^{x} \frac{\left(1-x\right)^{2}}{\left(1+x^{2}\right)^{2}}dx$
$\Rightarrow I = \int e^{x} \frac{1-x^{2} -2x}{\left(1+x^{2}\right)^{2}}dx$
$ = \int e^{x} \left[\frac{1-x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{2x}{\left(1+x^{2}\right)^{2}}\right]dx$
$\Rightarrow I = \int e^{x} \frac{1}{\left(1+x^{2}\right)}dx-2 \int e^{x}. \frac{x}{\left(1+x^{2}\right)^{2}}dx$
$\Rightarrow I =\frac{1}{\left(1+x^{2}\right)}.e^{x}-\int e^{x}. \left[-\left(1+x^{2}\right)^{-2}\right]2xdx-2\int\frac{e^{x}x}{\left(1+x^{2}\right)^{2}}dx$
$\Rightarrow I =\frac{e^{x}}{1+x^{2}}+2\int\frac{e^{x}x}{\left(1+x^{2}\right)^{2}}dx-2\int\frac{e^{x}x}{\left(1+x^{2}\right)^{2}}dx$
$\Rightarrow I =\frac{e^{x}}{1+x^{2}}+C$