Question

If $\int x\frac{\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=a\sqrt{1+x^2}\ln (x+\sqrt{1+x^2})+bx+c$, then

• A. $a=1,b=-1$
• B. $a=1,b=1$
• C. $a=-1,b=1$
• D. $a=-1,b=-1$

Answer 1

The correct option is A $a=1,b=-1$

$I=\int x\frac{\ln (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}dx$
Let $t=\sqrt{x^2+1}$
or $\frac{dt}{dx}=\frac{x}{\sqrt{x^2+1}}$
$\therefore I=\int \ln (t+\sqrt{t^2+1})dt$
$=t\ln (t+\sqrt{t^2-1})-\int \frac{1+\frac{t}{\sqrt{t^2-1}}}{t+\sqrt{t^2-1}}tdt$
$=t\ln (t+\sqrt{t^2-1})-\frac{1}{2}\int \frac{2t}{\sqrt{t^2-1}}dt$
$=t\ln (t+\sqrt{t^2-1})-\sqrt{t^2-1}$
Resubstituting the value of $x$,
$=\sqrt{1+x^2}\ln (x+\sqrt{1+x^2})-x+c$
On comparing we get $a=1,b=-1$