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

The integral of (x^2)/((x+1)(x-1)^2) is 3/4 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 (x^2)/((x+1)(x-1)^2)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{( u + 1 ) ^{2}}{u ^{2} ( u + 2 )} : \frac{3}{4 u} + \frac{1}{2 u ^{2}} + \frac{1}{4 ( u + 2 )}

= \int \frac{3}{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{3}{4 u} d u + \int \frac{1}{2 u ^{2}} d u + \int \frac{1}{4 ( u + 2 )} d u

\int \frac{3}{4 u} d u = \frac{3}{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{3}{4} ln ∣ u ∣ - \frac{1}{2 u} + \frac{1}{4} ln ∣ u + 2 ∣

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{2} x^{- 3} d x

t y^{^{'}} + (t + 1) y = t

\int 6 x (3 x^{2} - 2)^{4} d x

\int_{0}^{\frac{1}{5}} 4 arcsin (5 y) d y

y = 3 x - 4 x^{2}, (- 2, - 22)

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