Question

$\int \frac{e^{x^2}\left(2x+x^3\right)}{{\left(3+x^2\right)}^2}dx$ is equal to :

• A. $\frac{e^{x^2}}{\left(3+x^2\right)}+k$
• B. $\frac{1}{2}\frac{e^{x^2}}{{\left(3+x^2\right)}^2}+k$
• C. $\frac{1}{4}\frac{e^{x^2}}{{\left(3+x^2\right)}^2}+k$
• D. $\frac{1}{2}\frac{e^{x^2}}{\left(3+x^2\right)}+k$

Answer 1

Answer: (D)

Put $x^2=t\Rightarrow 2xdx=dt$
$I=\int \frac{e^{x^2}\left(2+x^2\right)xdx}{{\left(3+x^2\right)}^2}=\frac{1}{2}\int e^t\frac{(2+t)}{(3+t{)}^2}dt$
$=\frac{1}{2}\int \frac{e^t(3+t-1)}{(3+t{)}^2}dt=\frac{1}{2}\int e^t\left[\frac{1}{3+t}-\frac{1}{(3+t{)}^2}\right]dt$
$=\frac{1}{2}{e}^t\frac{1}{3+t}+k\left[\because \frac{d}{dt}\left(\frac{1}{3+t}\right)=\frac{-1}{(3+t{)}^2}\right]$
$=\frac{1}{2}\frac{e^{x^2}}{3+x^2}+k$