Question

$\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ is equal to

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

Answer 1

Answer: (D)

Let $I=\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$
Put $t=\log (x+\sqrt{1+x^2})$ $\frac{dt}{dx}=\frac{1}{x+\sqrt{1+x^2}}\left\{1+\frac{x}{\sqrt{1+x^2}}\right\}$
$=\frac{x+\sqrt{1+x^2}}{(x+\sqrt{1+x^2}).\sqrt{1+x^2}}$
$=\frac{1}{\sqrt{1+x^2}}$
$\therefore$ $\frac{dx}{\sqrt{1+x^2}}=dt$
$\therefore$ $I=\int tdt=\frac{t^2}{2}+c$
$=\frac{1}{2}{[\log (x+\sqrt{1+x^2})]}^2+c$