Question

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

Answer 1

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`