How do you integrate $ln(x+sqrt(x^2+1))$ ?

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Answer 1 · 2016-06-05T20:30:33

By parts.

You can integrate it by parts with the rule

$\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)$

where we assume that

$f'(x)=1$ and $g(x)=ln(x+sqrt(x^2+1))$

consequently

$f(x)=x$ and $g'(x)=1/(sqrt(x^2+1))$ .

The integral is then

$\int ln(x+sqrt(x^2+1))dx$

$=x ln(x+sqrt(x^2+1))-\int x/(sqrt(x^2+1))dx$

$=x ln(x+sqrt(x^2+1))-sqrt(x^2+1) + C$ .

How do you integrate ln(x+sqrt(x^2+1))ln(x+sqrt(x^2+1))?