What is the integral of 6/(3x^2-2x-8) ?

The integral of 6/(3x^2-2x-8) is 6(-1/10 ln|3x+4|+1/10 ln|x-2|)+C

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integral of 6/(3x^2-2x-8)

\int \frac{6}{3 x ^{2} - 2 x - 8} d x

6 (- \frac{1}{10} ln ∣ 3 x + 4 ∣ + \frac{1}{10} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{6}{3 x ^{2} - 2 x - 8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{3 x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{1}{3 x ^{2} - 2 x - 8} : - \frac{3}{10 ( 3 x + 4 )} + \frac{1}{10 ( x - 2 )}

= 6 \cdot \int - \frac{3}{10 ( 3 x + 4 )} + \frac{1}{10 ( x - 2 )} 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

= 6 (- \int \frac{3}{10 ( 3 x + 4 )} d x + \int \frac{1}{10 ( x - 2 )} d x)

\int \frac{3}{10 ( 3 x + 4 )} d x = \frac{1}{10} ln ∣ 3 x + 4 ∣

\int \frac{1}{10 ( x - 2 )} d x = \frac{1}{10} ln ∣ x - 2 ∣

= 6 (- \frac{1}{10} ln ∣ 3 x + 4 ∣ + \frac{1}{10} ln ∣ x - 2 ∣)

Add a constant to the solution

= 6 (- \frac{1}{10} ln ∣ 3 x + 4 ∣ + \frac{1}{10} ln ∣ x - 2 ∣) + C

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What is the integral of 6/(3x^2-2x-8) ?
The integral of 6/(3x^2-2x-8) is 6(-1/10 ln|3x+4|+1/10 ln|x-2|)+C