Question

Evaluate: $\int \frac{x}{\sqrt{1+x^2}}dx$

Answer 1

Put, $1+x^2=t⟹2xdx=dt$

$\int \frac{x}{\sqrt{1+x^2}}dx=\int \frac{1}{2\sqrt{t}}dt$

$=\frac{1}{2}\frac{(t{)}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$

$=\frac{1}{3}(1+x^2{)}^{\frac{3}{2}}+c$