What is the integral of x/((x+1)^4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/((x+1)^4)

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

- \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{u ^{4}} d u

= \int \frac{u - 1}{u ^{4}} d u

Expand

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} 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 ^{3}} d u - \int \frac{1}{u ^{4}} d u

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

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

= - \frac{1}{2 u ^{2}} - (- \frac{1}{3 u ^{3}})

Substitute back

u = x + 1

= - \frac{1}{2 ( x + 1 ) ^{2}} - (- \frac{1}{3 ( x + 1 ) ^{3}})

Simplify

= - \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + C

n = 1 \sum \infty 3 e^{- 9 n}

f (x) = \frac{1}{x ^{2}}, [2, 6]

(ln (x))^{4}

\int 2 + e^{2 x} + e^{- 2 x} d x

\frac{\partial}{\partial y} (x^{3} - y^{3} + 2 x y)

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