How do you integrate #int (x+5)/(2x+3)# using substitution?

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Answer 1 · 2016-08-23T12:02:19

#=7/4ln(2x+3) + 1/2x + C#

We can't immediately substitute into this integrand. First we have to get it into a more receptive form:

We do this with polynomial long division. It's a very simple thing to do on paper but the formatting is quite difficult on here.

#int (x+5)/(2x+3)dx = int (7/(2(2x+3)) + 1/2)dx#

#=7/2int (dx)/(2x+3) + 1/2intdx#

Now for the first integral set #u = 2x+3 implies du = 2dx#

#implies dx = (du)/2#

#=7/4int(du)/(u) + 1/2intdx#

#=7/4ln(u) + 1/2x + C#

#=7/4ln(2x+3) + 1/2x + C#