How do you find the integral of #(4x)/(4x+7)dx#?

1 Answer
Jun 17, 2015

I like rewriting: #(4x)/(4x+7) = (4x+7-7)/(4x+7) = 1- 7/(4x+7)#

Explanation:

#int (4x)/(4x+7) = int( 1- 7/(4x+7)) dx#

#= x-7/4 ln abs(4x+7) +C#

Note
To evaluate #int 7/(4x+7) dx# use substitution with #u= 4x+7# so #dx = 1/4 du# and we have #7/4 int 1/u du#

Second Note

This integral can also be evaluated by integration by parts, with #u=x# and #dv = 4/(4x+7)#.

Parts gives us:

#xln abs(4x+7) - int ln (4x+7) dx#

The integral here can be found by substitution if you know #int lnu du# and by substitution and integration by parts if you don't know #int ln u du#.