Practicing Success
Practicing Success
CUET Preparation Today
The value of $\int\frac{\log[x+\sqrt{1+x^2}]}{\sqrt{1+x^2}}dx$ is: |
$[\log(x+\sqrt{1+x^2})]^2$ $\frac{1}{4}[\log(x+\sqrt{1+x^2})]$ $\frac{1}{2}[\log(x+\sqrt{1+x^2})]^2$ None of these |
$\frac{1}{2}[\log(x+\sqrt{1+x^2})]^2$ |
We know that $\int\frac{1}{\sqrt{1+x^2}}dx=\log(x+\sqrt{1+x^2})$ $∴\frac{d}{dx}\log(x+\sqrt{1+x^2})=\frac{1}{\sqrt{(1+x^2)}}$ $∴I=\int t\,dt=\frac{1}{2}t^2=\frac{1}{2}[\log(x+\sqrt{1+x^2})]^2$ |
