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#.