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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

cos (2 arctan (x))

\int sin (5 x + \frac{π}{6}) d x

\frac{1}{2} x^{2} + 8

y = π r^{2}

(e^{x} + x)^{4}

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