What is the integral of (3x+5)/((x+1)(x-1)^2) ?

The integral of (3x+5)/((x+1)(x-1)^2) is 1/2 ln|x+1|-1/2 ln|x-1|-4/(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 (3x+5)/((x+1)(x-1)^2)

\int \frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}} d x

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

Solution steps

\int \frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}} d x

Take the partial fraction of

\frac{3 x + 5}{( x + 1 ) ( x - 1 ) ^{2}} : \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} + \frac{4}{( x - 1 ) ^{2}}

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

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

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

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

\int \frac{1}{20 + x ^{2}} d x

\int (sin (x) - 1)^{4} cos (x) d x

\int y^{4} ln (y) d y

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

\int (0.1 q + 1) e^{0.3 q} d q

What is the integral of (3x+5)/((x+1)(x-1)^2) ?
The integral of (3x+5)/((x+1)(x-1)^2) is 1/2 ln|x+1|-1/2 ln|x-1|-4/(x-1)+C