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

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

integral

\frac{x - 1}{( x + 1 ) ^{3}}

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{u ^{3}} d u

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

Expand

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

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\frac{cos ^{2} ( t )}{2}

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

\frac{x}{( x + 1 )}

3 - 2 x

\frac{2}{( 3 + e ^{x} )}

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