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

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

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

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

\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1) + C

Solution steps

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

Complete the square

(x - 2) (x + 4) : (x + 1)^{2} - 9

= \int \frac{x}{( x + 1 ) ^{2} - 9} d x

Apply u-substitution

= \int \frac{u - 1}{u ^{2} - 9} d u

Expand

\frac{u - 1}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} - \frac{1}{u ^{2} - 9}

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{u}{u ^{2} - 9} d u - \int \frac{1}{u ^{2} - 9} d u

\int \frac{u}{u ^{2} - 9} d u = \frac{1}{2} ln u^{2} - 9

\int \frac{1}{u ^{2} - 9} d u = - \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

= \frac{1}{2} ln u^{2} - 9 - (- \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1))

Substitute back

u = x + 1

= \frac{1}{2} ln (x + 1)^{2} - 9 - (- \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1))

Simplify

\frac{1}{2} ln (x + 1)^{2} - 9 - (- \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1)) : \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)

= \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1) + C

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