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

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Answer 1 · 2018-04-09T18:46:36

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

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#