How do you integrate #x/(x+10)#?

1 Answer
Apr 9, 2018

#intx/(x+10)dx=x-10ln|x+10|+C#

Explanation:

A simple substitution will do.

#u=x+10 -> x=u-10#

#du=dx#

Rewrite and simplify:

#int(u-10)/udu=intu/udu-10int(du)/u#

Integrate:

#intdu-10int(du)/u=u-10ln|u|+C#

Rewrite in terms of #x,# yielding

#intx/(x+10)dx=x+10-10ln|x+10|+C#

We may absorb the #10# into #C.#

#intx/(x+10)dx=x-10ln|x+10|+C#