Question

$\int x \cdot \frac{\ln \left( x +\sqrt{1+ x ^2}\right)}{\sqrt{1+ x ^2}} dx$ equals :

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

Answer 1

Answer: (A)

use I.B.P. taking $\ln \left(x+\sqrt{1+x^2}\right)$ as the first function and $\frac{x}{\sqrt{1+x^2}}$ as the second function