Practicing Success
Practicing Success
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$\int \frac{\left\{x+\sqrt{x^2+1}\right\}^n}{\sqrt{x^2+1}} d x$ is equal to |
$\left\{x+\sqrt{x^2+1}\right\}^n+C$ $\frac{1}{n}\left\{x+\sqrt{x^2+1}\right\}^n+C$ $\frac{1}{n+1}\left\{x+\sqrt{x^2+1}\right\}^{n+1}+C$ none of these |
$\frac{1}{n}\left\{x+\sqrt{x^2+1}\right\}^n+C$ |
Let $I=\int \frac{\left\{x+\sqrt{x^2+1}\right\}^n}{\sqrt{x^2+1}} d x$ $\Rightarrow I=\int\left\{x+\sqrt{x^2+1}\right\}^{n-1} \times \frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}} d x$ $\Rightarrow I=\int\left\{x+\sqrt{x^2+1}\right\}^{n-1} d\left(x+\sqrt{x^2+1}\right)$ $\Rightarrow I=\frac{1}{n}\left(x+\sqrt{x^2+1}\right)^n+C$ |
