The value of $\int\frac{\log[x+\sqrt{1+x^2}]}{\sqrt{1+x^2}}dx$ is:
Answer & explanation
Correct answer: option 3
We know that $\int\frac{1}{\sqrt{1+x^2}}dx=\log(x+\sqrt{1+x^2})$
$∴\frac{d}{dx}\log(x+\sqrt{1+x^2})=\frac{1}{\sqrt{(1+x^2)}}$
$∴I=\int t\,dt=\frac{1}{2}t^2=\frac{1}{2}[\log(x+\sqrt{1+x^2})]^2$