How do you use substitution to integrate #(5x+10)/(3x^(2) +12x - 7)#?

1 Answer
Jim H
Sep 6, 2015

Start by factoring the 5 out of the numerator.

Explanation:

#int (5x+10)/(3x^(2) +12x - 7) dx = 5int (x+2)/(3x^2+12x-7) dx#

Now the derivative of the denominator is #6# times the numerator, so do a substitution to get a natural logarithm.

Let #u = 3x^2+12x-7# which makes #du = (6x+12) dx#

So our integral becomes: #5/6 int 1/u du = 5/6 ln abs u +C#.

And reversing the substitution gets us

#int (5x+10)/(3x^(2) +12x - 7) dx = 5/6 ln abs (3x^2+12x-7) +C#

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