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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/((x-1)^3)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{( x - 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{u + 1}{u ^{3}} d u

= 2 \cdot \int \frac{u + 1}{u ^{3}} d u

Expand

\frac{u + 1}{u ^{3}} : \frac{1}{u ^{2}} + \frac{1}{u ^{3}}

= 2 \cdot \int \frac{1}{u ^{2}} + \frac{1}{u ^{3}} 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

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

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

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= 2 (- \frac{1}{u} - \frac{1}{2 u ^{2}})

Substitute back

u = x - 1

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

Simplify

2 (- \frac{1}{x - 1} - \frac{1}{2 ( x - 1 ) ^{2}}) : - \frac{2}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

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

Add a constant to the solution

= - \frac{2}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

\int \frac{[ cos ( θ - 1 ) ]}{[ sin ^{2} ( θ - 1 ) ]} d θ

\int \frac{5 x + 31}{3 x ^{2} - 4 x + 11} d x

\int 2 (2 x + 1)^{9} d x

\int \frac{e ^{18 x}}{1 + e ^{6 x}} d x

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

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