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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u ^{2} + 2 u )} d u

= \int \frac{1}{u ( u ^{2} + 2 u )} d u

Take the partial fraction of

\frac{1}{u ( u ^{2} + 2 u )} : - \frac{1}{4 u} + \frac{1}{2 u ^{2}} + \frac{1}{4 ( u + 2 )}

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

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

= - \frac{1}{4} ln ∣ u ∣ - \frac{1}{2 u} + \frac{1}{4} ln ∣ u + 2 ∣

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int (u^{\frac{3}{2}} - 3 u + 14) d u

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

\int [6 e^{u} - u^{3} (u + 1)] d u

\int 9 x x - 7 d x

\int 8 \cdot 2^{- x} - 4 d x

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