Question

$\int \frac{x}{(1+x^2{)}^{3/2}}dx=$

• A. $\frac{-1}{\sqrt{1+x^2}}+c$
• B. $\frac{1}{(1+x^2)}+c$
• C. $\frac{-1}{(1+x^2{)}^{3/2}}+c$
• D. $\frac{-1}{(1+x^2)}+c$

Answer 1

The correct option is A $\frac{-1}{\sqrt{1+x^2}}+c$

$I=\int \frac{x}{(1+x^2{)}^{3/2}}dx$

$x^2=t$

$2xdx=dt$

$xdx=\frac{1}{2}dt$
Then,

$I=\int \frac{1}{2}\frac{dt}{(1+t{)}^{3/2}}=\frac{1}{2}\int \frac{dt}{(1+t{)}^{3/2}}$

$=1/2(-2)(1+t{)}^{1/2}+c$

$=-\frac{1}{(1+t{)}^{1/2}}+c$

$=-\frac{1}{\sqrt{1+x^2}}+c(\because t=x^2)$

$\therefore \int \frac{x}{(1+x^2{)}^{3/2}}=\frac{-1}{\sqrt{1+x^2}}+c$