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

Question

1127 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T22:05:17

#int ln(x^2+1) dx = xln(x^2+1)- 2 x+2arctanx+C#

Integrate by parts:

#int ln(x^2+1) dx = int ln(x^2+1) d/dx (x) dx#

#int ln(x^2+1) dx = xln(x^2+1)- int x d/dx( ln(x^2+1) ) dx#

#int ln(x^2+1) dx = xln(x^2+1)- 2 int x^2/(x^2+1) dx#

Now:

#int ln(x^2+1) dx = xln(x^2+1)- 2 int (x^2+1-1)/(x^2+1) dx#

and using the linearity of the integral:

#int ln(x^2+1) dx = xln(x^2+1)- 2 int dx+2int 1/(x^2+1) dx#

These are well known integrals, so:

#int ln(x^2+1) dx = xln(x^2+1)- 2 x+2arctanx+C#