What is the integral of (x+10)/((x+4)(x-2)) ?

The integral of (x+10)/((x+4)(x-2)) is -ln|x+4|+2ln|x-2|+C

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integral of (x+10)/((x+4)(x-2))

\int \frac{x + 10}{( x + 4 ) ( x - 2 )} d x

- ln ∣ x + 4 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x + 10}{( x + 4 ) ( x - 2 )} d x

Apply u-substitution

: \int \frac{u + 12}{u ( u + 6 )} d u

= \int \frac{u + 12}{u ( u + 6 )} d u

Expand

\frac{u + 12}{u ( u + 6 )} : \frac{u}{u ( u + 6 )} + \frac{12}{u ( u + 6 )}

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 ( u + 6 )} d u + \int \frac{12}{u ( u + 6 )} d u

\int \frac{u}{u ( u + 6 )} d u = ln ∣ u + 6 ∣

\int \frac{12}{u ( u + 6 )} d u = 12 (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

= ln ∣ u + 6 ∣ + 12 (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

Substitute back

u = x - 2

= ln ∣ x - 2 + 6 ∣ + 12 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x - 2 + 6 ∣)

Simplify

ln ∣ x - 2 + 6 ∣ + 12 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x - 2 + 6 ∣) : - ln ∣ x + 4 ∣ + 2 ln ∣ x - 2 ∣

= - ln ∣ x + 4 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x + 4 ∣ + 2 ln ∣ x - 2 ∣ + C

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

\int 4 x - 6 x^{2} + 3 + 2 sin (x) - 3 e^{x} + \frac{1}{x} + x^{- 3} d x

\int 7 x x + 5 d x

\int \frac{1}{2 cos ^{2} ( 3 x )} d x

\int x^{\frac{1}{10}} d x

What is the integral of (x+10)/((x+4)(x-2)) ?
The integral of (x+10)/((x+4)(x-2)) is -ln|x+4|+2ln|x-2|+C