What is the integral of (12x^2+4x-16)/(6x^2-13x-5) ?

The integral of (12x^2+4x-16)/(6x^2-13x-5) is 2x+16/17 ln|3x+1|+69/17 ln|2x-5|+C

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integral of (12x^2+4x-16)/(6x^2-13x-5)

\int \frac{12 x ^{2} + 4 x - 16}{6 x ^{2} - 13 x - 5} d x

2 x + \frac{16}{17} ln ∣ 3 x + 1 ∣ + \frac{69}{17} ln ∣ 2 x - 5 ∣ + C

Solution steps

\int \frac{12 x ^{2} + 4 x - 16}{6 x ^{2} - 13 x - 5} d x

Take the partial fraction of

\frac{12 x ^{2} + 4 x - 16}{6 x ^{2} - 13 x - 5} : 2 + \frac{48}{17 ( 3 x + 1 )} + \frac{138}{17 ( 2 x - 5 )}

= \int 2 + \frac{48}{17 ( 3 x + 1 )} + \frac{138}{17 ( 2 x - 5 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 d x + \int \frac{48}{17 ( 3 x + 1 )} d x + \int \frac{138}{17 ( 2 x - 5 )} d x

\int 2 d x = 2 x

\int \frac{48}{17 ( 3 x + 1 )} d x = \frac{16}{17} ln ∣ 3 x + 1 ∣

\int \frac{138}{17 ( 2 x - 5 )} d x = \frac{69}{17} ln ∣ 2 x - 5 ∣

= 2 x + \frac{16}{17} ln ∣ 3 x + 1 ∣ + \frac{69}{17} ln ∣ 2 x - 5 ∣

Add a constant to the solution

= 2 x + \frac{16}{17} ln ∣ 3 x + 1 ∣ + \frac{69}{17} ln ∣ 2 x - 5 ∣ + C

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What is the integral of (12x^2+4x-16)/(6x^2-13x-5) ?
The integral of (12x^2+4x-16)/(6x^2-13x-5) is 2x+16/17 ln|3x+1|+69/17 ln|2x-5|+C