Question

Integral of $f\left(x\right)=\sqrt{1+x^2}$ with respect to $x^2$ is

• A. $\frac{2}{3}\frac{{(1+x^2)}^{\frac{3}{2}}}{x}+k$
• B. $\frac{2}{3}{(1+x^2)}^{\frac{3}{2}}+k$
• C. $\frac{2}{3}x{(1+x^2)}^{\frac{3}{2}}+k$
• D. none of these

Answer 1

Answer: (B)

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

Let $I=\int f\left(x\right)d\left(x^2\right)=\int 2x\sqrt{1+x^2}dx$
Substitute $t=\sqrt{1+x^2}\Rightarrow dt=2xdx$
$\therefore I=\int \sqrt{t}dt=\frac{2}{3}{t}^{\frac{3}{2}}+c$
$=\frac{2}{3}{(1+x^2)}^{\frac{3}{2}}+k$