What is the integral of (x-5)/((x+1)(x-2)) ?

The integral of (x-5)/((x+1)(x-2)) is -ln|x-2|+2ln|x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-5)/((x+1)(x-2))

\int \frac{x - 5}{( x + 1 ) ( x - 2 )} d x

- ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x - 5}{( x + 1 ) ( x - 2 )} d x

Apply u-substitution

: \int \frac{u - 6}{u ( u - 3 )} d u

= \int \frac{u - 6}{u ( u - 3 )} d u

Expand

\frac{u - 6}{u ( u - 3 )} : \frac{u}{u ( u - 3 )} - \frac{6}{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{6}{u ( u - 3 )} d u

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

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

= ln ∣ u - 3 ∣ - 6 (- \frac{1}{3} ln ∣ u ∣ + \frac{1}{3} ln ∣ u - 3 ∣)

Substitute back

u = x + 1

= ln ∣ x + 1 - 3 ∣ - 6 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣)

Simplify

ln ∣ x + 1 - 3 ∣ - 6 (- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x + 1 - 3 ∣) : - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣

= - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x - 2 ∣ + 2 ln ∣ x + 1 ∣ + C

\int tan^{10} (x) sec^{4} (x) d x

\int \frac{8}{sec ( x ) - 1} d x

\int \frac{1}{x ( x ^{100} + 1 )} d x

\int (4 x + 3) 1 + 2 x d x

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

What is the integral of (x-5)/((x+1)(x-2)) ?
The integral of (x-5)/((x+1)(x-2)) is -ln|x-2|+2ln|x+1|+C