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$\Large \int e^x\bigg(\frac{1-x}{1+x^2}\bigg)^2 \normalsize dx$ is equal to

$\begin{array}{1 1} (A)\;\frac{e^x}{1+x^2}+C & (B)\;\frac{-e^x}{1+x^2}+C\\(C)\;\frac{e^x}{(1+x^2)^2}+C & (D)\;\frac{-e^x}{(1+x^2)^2}+C \end{array} $

1 Answer

Toolbox:
  • $\int e^x[f(x)+f'(x)]=e^xf(x)$
  • $\large\frac{d}{dx}\big(\frac{1}{x}\big)=\frac{-1}{x^2}$
Step 1:
Let $I=\int e^x\bigg(\large\frac{1-x}{1+x^2}\bigg)^2$$dx$
$(1-x)^2=1-2x+x^2=1+x^2-2x$
$\quad=\int e^x\bigg(\large\frac{1+x^2-2x}{(1+x^2)^2}\bigg)$$dx$
On splitting the terms,
$\quad=\int e^x\begin{bmatrix}\large\frac{1}{(1+x^2)}-\frac{2x}{(1+x^2)^2}\end{bmatrix}$$dx$
Step 2:
If $f(x)=\large\frac{1}{1+x^2}$ then $f'(x)=\large\frac{-2x}{(1+x^2)^2}$
Clearly this is of the form
$\int e^x[f(x)+f'(x)dx]=e^xf(x)+c$
Hence $I=\large\frac{e^x}{1+x^2}$$+c$.
Hence the correct option is A.
answered Apr 24, 2013 by sreemathi.v
 

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