What is the integral of x/((x+1)^8) ?

The integral of x/((x+1)^8) is -1/(6(x+1)^6)+1/(7(x+1)^7)+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)^8)

\int \frac{x}{( x + 1 ) ^{8}} d x

- \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}} + C

Solution steps

\int \frac{x}{( x + 1 ) ^{8}} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{8}} d u

= \int \frac{u - 1}{u ^{8}} d u

Expand

\frac{u - 1}{u ^{8}} : \frac{1}{u ^{7}} - \frac{1}{u ^{8}}

= \int \frac{1}{u ^{7}} - \frac{1}{u ^{8}} 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 ^{7}} d u - \int \frac{1}{u ^{8}} d u

\int \frac{1}{u ^{7}} d u = - \frac{1}{6 u ^{6}}

\int \frac{1}{u ^{8}} d u = - \frac{1}{7 u ^{7}}

= - \frac{1}{6 u ^{6}} - (- \frac{1}{7 u ^{7}})

Substitute back

u = x + 1

= - \frac{1}{6 ( x + 1 ) ^{6}} - (- \frac{1}{7 ( x + 1 ) ^{7}})

Simplify

= - \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}}

Add a constant to the solution

= - \frac{1}{6 ( x + 1 ) ^{6}} + \frac{1}{7 ( x + 1 ) ^{7}} + C

\int \frac{e ^{5 x}}{e ^{5 x} + 4} d x

\frac{\partial}{\partial y} (π x^{2} y)

\int 21 x^{2} ln (x) d x

\frac{d y}{d t} = t e^{2 y}

\int x ln (3 x^{2}) d x

What is the integral of x/((x+1)^8) ?
The integral of x/((x+1)^8) is -1/(6(x+1)^6)+1/(7(x+1)^7)+C