$\int \frac{\left\{x+\sqrt{x^2+1}\right\}^n}{\sqrt{x^2+1}} d x$ is equal to
Answer & explanation
Correct answer: option 2
Let
$I=\int \frac{\left\{x+\sqrt{x^2+1}\right\}^n}{\sqrt{x^2+1}} d x$
$\Rightarrow I=\int\left\{x+\sqrt{x^2+1}\right\}^{n-1} \times \frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}} d x$
$\Rightarrow I=\int\left\{x+\sqrt{x^2+1}\right\}^{n-1} d\left(x+\sqrt{x^2+1}\right)$
$\Rightarrow I=\frac{1}{n}\left(x+\sqrt{x^2+1}\right)^n+C$