Question

$\int x\frac{\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ equals to

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

Answer 1

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

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

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

Let

$I_1=\int \frac{2x}{2\sqrt{1+x^2}}.$

Put $1+x^2=t⟹2xdx=dt$

$I_1=\int \frac{1}{2}{t}^{\frac{-1}{2}}dt=t^{\frac{1}{2}}=\sqrt{1+x^2}$

Using integral by parts,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I=\ln (x+\sqrt{1+x^2})\int \frac{2x}{2\sqrt{1+x^2}}dx-\int [\frac{d\left\{\ln \right(x+\sqrt{1+x^2}\left)\right\}}{dx}.\int \frac{2x}{2\sqrt{1+x^2}}dx]dx+c$

$=\ln (x+\sqrt{1+x^2}).\sqrt{1+x^2}-\int [\frac{1}{x+\sqrt{1+x^2}}.(1+\frac{x}{\sqrt{1+x^2}}).\sqrt{1+x^2}]dx+c$

$=\ln (x+\sqrt{1+x^2}).\sqrt{1+x^2}-\int dx+c=\ln (x+\sqrt{1+x^2}).\sqrt{1+x^2}-x+c$

$=\sqrt{1+x^2}\ln (x+\sqrt{1+x^2})-x+c$