Question

$\int x\frac{\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ equals
$($where $C$ is the constant of integration$)$

• A. $\sqrt{1+x^2}\cdot \ln (x+\sqrt{1+x^2})-x+C$
• B. $\sqrt{1+x^2}\cdot \ln (x+\sqrt{1+x^2})+x+C$
• C. $\frac{x}{2}\cdot {\ln }^2(x+\sqrt{1+x^2})-\frac{x}{\sqrt{1+x^2}}+C$
• D. $\frac{x}{2}\cdot \ln (x+\sqrt{1+x^2})+\frac{x}{\sqrt{1+x^2}}+C$

Answer 1

The correct option is A $\sqrt{1+x^2}\cdot \ln (x+\sqrt{1+x^2})-x+C$

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

Let $x+\sqrt{1+x^2}=t$ $\Rightarrow \sqrt{1+x^2}=t-x$
$\Rightarrow 1+x^2=t^2+x^2-2tx$
$\Rightarrow x=\frac{t^2-1}{2t}\dots \left(1\right)$
And $(1+\frac{2x}{2\sqrt{1+x^2}})dx=dt$
$\Rightarrow \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}dx=dt$
$\Rightarrow \frac{t}{\sqrt{1+x^2}}dx=dt$
$\Rightarrow \frac{dx}{\sqrt{1+x^2}}=\frac{dt}{t}$

$I=\int \frac{t^2-1}{2t}\cdot \frac{\left(\ln t\right)}{t}dt=\underset{I_1}{\underset{}{\frac{1}{2}\int \ln tdt}}-\underset{I_2}{\underset{}{\frac{1}{2}\int \frac{\ln t}{t^2}dt}}$

$I_1=\frac{1}{2}\int \ln tdt$
Applying by parts, we get
$I_1=\frac{1}{2}(t\ln t-t)+C_1$

$I_2=\frac{1}{2}\int \ln t\cdot \frac{1}{t^2}dt$
Again applying by parts
$=\frac{1}{2}\left[\ln t\int \frac{dt}{t^2}-\int (\frac{d\left(\ln t\right)}{dt}\int \frac{dt}{t^2})dt\right]$
$=\frac{1}{2}\left[(\ln t\left(\frac{-1}{t}\right)-\int \frac{-1}{t^2}dt)\right]$
$I_2=\frac{1}{2}[-\frac{\ln t}{t}-\frac{1}{t}]+C_2$

$\Rightarrow I=\frac{1}{2}(t\ln t-t+\frac{\ln t}{t}+\frac{1}{t})+C$
$=(t+\frac{1}{t})\ln t+\frac{1}{t}-t+C$
From equation $\left(1\right),$ $\frac{1}{t}-t=-2x$
and $\frac{1}{t}+t=\sqrt{{(\frac{1}{t}-t)}^2+4}=2\sqrt{x^2+1}$

$I=\frac{1}{2}\left[2\sqrt{1+x^2}\ln \right(x+\sqrt{1+x^2})-2x+C]$
$\therefore I=\sqrt{1+x^2}\cdot \ln (x+\sqrt{1+x^2})-x+C$