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

1 Answer
mason m
Jan 14, 2017

#3x+11ln(abs(x-2))+C#

Explanation:

First simplify the integrand. This can be done by performing long division, or by my preferred method:

#int(3x+5)/(x-2)dx=int(3(x-2)+11)/(x-2)dx#

#color(white)(int(3x+5)/(x-2)dx)=int(3(x-2))/(x-2)dx+int11/(x-2)dx#

#color(white)(int(3x+5)/(x-2)dx)=3intdx+11intdx/(x-2)#

Both of these are simple integration problems. If you're stuck on the second one, try the substitution #u=x-2=>du=dx# and recall that #int(du)/u=ln(absu)+C#.

#color(white)(int(3x+5)/(x-2)dx)=3x+11ln(abs(x-2))+C#

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