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

1 Answer
Jun 5, 2016

By parts.

Explanation:

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#.