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

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Answer 1 · 2015-09-06T17:25:36

Start by factoring the 5 out of the numerator.

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

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

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

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