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

The Correct Option is D

The correct answer is D:\(\frac{1}{2}\frac{e^{x^{2}}}{(3+x^2)}+k\)
Put \(x^{2}=t \Rightarrow 2 x d x=d t\)
\(I=\int \frac{e^{x^{2}}\left(2+x^{2}\right) x d x}{\left(3+x^{2}\right)^{2}}=\frac{1}{2} \int e^{t} \frac{(2+t)}{(3+t)^{2}} d t \)
\(=\frac{1}{2} \int \frac{e^{t}(3+t-1)}{(3+t)^{2}} d t=\frac{1}{2} \int e^{t}\left[\frac{1}{3+t}-\frac{1}{(3+t)^{2}}\right] d t \)
\(=\frac{1}{2} e^{t} \frac{1}{3+t}+k\left[\because \frac{d}{d t}\left(\frac{1}{3+t}\right)=\frac{-1}{(3+t)^{2}}\right] \)
\(=\frac{1}{2} \frac{e^{x^{2}}}{3+x^{2}}+k\)