Question

The value of integral $\int \frac{{(\sqrt{1+x^2}+x)}^n}{\sqrt{1+x^2}}dx$, is

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

Answer 1

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

Let, ${(\sqrt{1+x^2}+x)}^n=z$

Differentiating both side, we get

$\Rightarrow n{(\sqrt{1+x^2}+x)}^{n-1}(\frac{x}{\sqrt{1+x^2}}+1)dx=dz$

$\Rightarrow \frac{{(\sqrt{1+x^2}+x)}^n}{\sqrt{1+x^2}}dx=\frac{dz}{n}$

$\therefore$ Given integral $=\int \frac{dz}{n}=\frac{1}{n}z+c$

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

$\therefore \int \frac{{(\sqrt{1+x^2}+x)}^n}{\sqrt{1+x^2}}dx=\frac{1}{n}{(\sqrt{1+x^2}+x)}^n+c$