What is the integral of 8/((x-9)^3) ?

The integral of 8/((x-9)^3) is -4/((x-9)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 8/((x-9)^3)

\int \frac{8}{( x - 9 ) ^{3}} d x

- \frac{4}{( x - 9 ) ^{2}} + C

Solution steps

\int \frac{8}{( x - 9 ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{( x - 9 ) ^{3}} d x

Apply u-substitution

= 8 \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= 8 (- \frac{1}{2 u ^{2}})

Substitute back

u = x - 9

= 8 (- \frac{1}{2 ( x - 9 ) ^{2}})

Simplify

8 (- \frac{1}{2 ( x - 9 ) ^{2}}) : - \frac{4}{( x - 9 ) ^{2}}

= - \frac{4}{( x - 9 ) ^{2}}

Add a constant to the solution

= - \frac{4}{( x - 9 ) ^{2}} + C

\int y tan (x) d x

\int \frac{y}{y} d y

\int 3 + x e^{x - 1} d x

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

\int \frac{1}{4 - 25 x ^{2}} d x

What is the integral of 8/((x-9)^3) ?
The integral of 8/((x-9)^3) is -4/((x-9)^2)+C