How do you integrate #(2x + 3)/((x + 7) dx#?

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Answer 1 · 2015-06-29T16:15:46

You can cancel things out.

#int(2x + 3)/(x+7)dx#

#= int(2x + 3 + 11 - 11)/(x+7)dx#

#= int(2x + 14 - 11)/(x+7)dx#

#= int(2x + 14)/(x+7) - 11/(x+7)dx#

#= int(2cancel((x + 7)))/cancel((x+7)) - 11/(x+7)dx#

#= int2dx - 11int1/(x+7)dx#

#= color(blue)(2x - 11ln|x+7| + C)#