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

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

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

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

ln ∣ x + 4 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{u ( u + 3 )} d u

= \int \frac{u - 2}{u ( u + 3 )} d u

Expand

\frac{u - 2}{u ( u + 3 )} : \frac{u}{u ( u + 3 )} - \frac{2}{u ( u + 3 )}

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 + 3 )} d u - \int \frac{2}{u ( u + 3 )} d u

\int \frac{u}{u ( u + 3 )} d u = ln ∣ u + 3 ∣

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

= ln ∣ u + 3 ∣ - 2 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = x + 1

= ln ∣ x + 1 + 3 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 1 + 3 ∣)

Simplify

ln ∣ x + 1 + 3 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 1 + 3 ∣) : ln ∣ x + 4 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣)

= ln ∣ x + 4 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣)

Add a constant to the solution

= ln ∣ x + 4 ∣ - 2 (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x + 4 ∣) + C

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