Question

Integration by parts:
\[\int {\ln (x + \sqrt {1 + {x^2}} )dx} \]

Answer 1

\[x*\ln(x+\sqrt{1+x^2}) - \int \frac{1}{x + \sqrt{1+x^2}}*x *(1 + \frac{1}{\cancel2}*\frac{1}{\sqrt{1+x^2}}*\cancel{2}x)dx\]

Answer 2

Now only the second part.
\[\int \frac{ x *\sqrt{1+x^2 } + x^2}{ (x + \sqrt{1 +x^2})(\sqrt{1+x^2)}}dx\]

Answer 3

\[\int \frac{ x *\sqrt{1+x^2 } + x^2}{ x*\sqrt{1+x^2} + x^2 + 1}dx\]

Answer 4

....BRB..

Answer 5

I got it..

Answer 6

Your first step is good and the second part can be calculate :
\[\int {xd(\ln (x + \sqrt {1 + {x^2}} )) = \int {x\frac{1}
{{x + \sqrt {1 + {x^2}} }}(\frac{{x + \sqrt {1 + {x^2}} }}
{{\sqrt {1 + {x^2}} }})} } \]

Answer 7

Thanks a lot.