What is #int 4xln(x^2+1) dx#?

Question

1642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T09:27:05

#int 4xln(x^2+1)dx = 2 (x^2+1)ln(x^2+1)-2x^2 +C#

Substitute:

#u = x^2+1#

#du = 2xdx#

so:

#int 4xln(x^2+1)dx = 2 int ln u du#

Integrate now by parts:

#int ln u du = u ln u - int u d(lnu)#

#int ln u du = u ln u - int u (du)/u#

#int ln u du = u ln u - int du#

#int ln u du = u ln u -u +C#

#int ln u du = u (ln u -1) +C#

undoing the substitution:

#int 4xln(x^2+1)dx = 2 (x^2+1)(ln(x^2+1)-1) +C#

and simplifying:

#int 4xln(x^2+1)dx = 2 (x^2+1)ln(x^2+1)-2x^2 +C#

where we absorbed all constants in #C#.