What is the integral of (x+8)/(3x^2-12) ?

The integral of (x+8)/(3x^2-12) is -1/2 ln|x+2|+5/6 ln|x-2|+C

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integral of (x+8)/(3x^2-12)

\int \frac{x + 8}{3 x ^{2} - 12} d x

- \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x + 8}{3 x ^{2} - 12} d x

Take the partial fraction of

\frac{x + 8}{3 x ^{2} - 12} : - \frac{1}{2 ( x + 2 )} + \frac{5}{6 ( x - 2 )}

= \int - \frac{1}{2 ( x + 2 )} + \frac{5}{6 ( 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

= - \int \frac{1}{2 ( x + 2 )} d x + \int \frac{5}{6 ( x - 2 )} d x

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

\int \frac{5}{6 ( x - 2 )} d x = \frac{5}{6} ln ∣ x - 2 ∣

= - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 2 ∣ + C

\int (\frac{1}{x ( ln ( x ) - 1 )}) d x

\int \frac{lo g _{3} ( x )}{3 x} d x

\int sin (4 x) cos^{2} (2 x) sin (x) d x

tan (3) x^{2}

\frac{\partial}{\partial x} (y^{x^{2} - y})

What is the integral of (x+8)/(3x^2-12) ?
The integral of (x+8)/(3x^2-12) is -1/2 ln|x+2|+5/6 ln|x-2|+C