Question

Evaluate $\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx.$

• A. $\frac{{\left[\log (x+\sqrt{1+x^2})\right]}^2}{2}+c$
• B. $\frac{1}{2{\left[\log (x+\sqrt{1+x^2})\right]}^2}+c$
• C. $\frac{{\left[\log (x+\sqrt{1+x^2})\right]}^2}{4}+c$
• D. $\frac{{\left[\log (x-\sqrt{1+x^2})\right]}^2}{2}+c$

Answer 1

The correct option is A $\frac{{\left[\log (x+\sqrt{1+x^2})\right]}^2}{2}+c$

$I=\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx.$

Put $\log (x+\sqrt{1+x^2})=tAnd\frac{1}{\sqrt{1+x^2}}dx=dt.$

$\therefore I=\int tdt=\frac{t^2}{2}+c$

$I=\frac{{\left[\log (x+\sqrt{1+x^2})\right]}^2}{2}+c$

Hence option ${}^{\prime }{A}^{\prime }$ is the answer.