What is the integral of 1/((x+4)^2(x-2)) ?

The integral of 1/((x+4)^2(x-2)) is -1/36 ln|x+4|+1/(6(x+4))+1/36 ln|x-2|+C

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integral of 1/((x+4)^2(x-2))

\int \frac{1}{( x + 4 ) ^{2} ( x - 2 )} d x

- \frac{1}{36} ln ∣ x + 4 ∣ + \frac{1}{6 ( x + 4 )} + \frac{1}{36} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{( x + 4 ) ^{2} ( x - 2 )} d x

Take the partial fraction of

\frac{1}{( x + 4 ) ^{2} ( x - 2 )} : - \frac{1}{36 ( x + 4 )} - \frac{1}{6 ( x + 4 ) ^{2}} + \frac{1}{36 ( x - 2 )}

= \int - \frac{1}{36 ( x + 4 )} - \frac{1}{6 ( x + 4 ) ^{2}} + \frac{1}{36 ( 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}{36 ( x + 4 )} d x - \int \frac{1}{6 ( x + 4 ) ^{2}} d x + \int \frac{1}{36 ( x - 2 )} d x

\int \frac{1}{36 ( x + 4 )} d x = \frac{1}{36} ln ∣ x + 4 ∣

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

\int \frac{1}{36 ( x - 2 )} d x = \frac{1}{36} ln ∣ x - 2 ∣

= - \frac{1}{36} ln ∣ x + 4 ∣ - (- \frac{1}{6 ( x + 4 )}) + \frac{1}{36} ln ∣ x - 2 ∣

Simplify

= - \frac{1}{36} ln ∣ x + 4 ∣ + \frac{1}{6 ( x + 4 )} + \frac{1}{36} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{36} ln ∣ x + 4 ∣ + \frac{1}{6 ( x + 4 )} + \frac{1}{36} ln ∣ x - 2 ∣ + C

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What is the integral of 1/((x+4)^2(x-2)) ?
The integral of 1/((x+4)^2(x-2)) is -1/36 ln|x+4|+1/(6(x+4))+1/36 ln|x-2|+C