Question

Evaluate: $\int \frac{\log [x+\sqrt{1+x^2}]}{\sqrt{1+x^2}}dx$

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

Answer 1

Answer: (B)

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

Let $I=\int \frac{\log [x+\sqrt{1+x^2}]}{\sqrt{1+x^2}}dx$
Put $\log (x+\sqrt{1+x^2})=t$
$\Rightarrow \frac{x+\sqrt{1+x^2}}{(x+\sqrt{1+x^2})\sqrt{1+x^2}}dx=\frac{1}{\sqrt{1+x^2}}dx=dt$
Therefore
$I=\int tdt=\frac{1}{2}{t}^2=\frac{1}{2}{\left[\log (x+\sqrt{1+x^2})\right]}^2$
Hence, option 'B' is correct.