What is the integral of (2x)/((2-x)^6) ?

The integral of (2x)/((2-x)^6) is -2(1/(4(2-x)^4)-2/(5(2-x)^5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/((2-x)^6)

\int \frac{2 x}{( 2 - x ) ^{6}} d x

- 2 (\frac{1}{4 ( 2 - x ) ^{4}} - \frac{2}{5 ( 2 - x ) ^{5}}) + C

Solution steps

\int \frac{2 x}{( 2 - x ) ^{6}} 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}{( 2 - x ) ^{6}} d x

Apply u-substitution

: \int - \frac{- u + 2}{u ^{6}} d u

= 2 \cdot \int - \frac{- u + 2}{u ^{6}} d u

Take the constant out:

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

= 2 (- \int \frac{- u + 2}{u ^{6}} d u)

Expand

\frac{- u + 2}{u ^{6}} : - \frac{1}{u ^{5}} + \frac{2}{u ^{6}}

= 2 (- \int - \frac{1}{u ^{5}} + \frac{2}{u ^{6}} 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 ^{5}} d u + \int \frac{2}{u ^{6}} d u))

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{2}{u ^{6}} d u = - \frac{2}{5 u ^{5}}

= 2 (- (- (- \frac{1}{4 u ^{4}}) - \frac{2}{5 u ^{5}}))

Substitute back

u = 2 - x

= 2 (- (- (- \frac{1}{4 ( 2 - x ) ^{4}}) - \frac{2}{5 ( 2 - x ) ^{5}}))

Simplify

= - 2 (\frac{1}{4 ( 2 - x ) ^{4}} - \frac{2}{5 ( 2 - x ) ^{5}})

Add a constant to the solution

= - 2 (\frac{1}{4 ( 2 - x ) ^{4}} - \frac{2}{5 ( 2 - x ) ^{5}}) + C

9 π x^{2}

\int \frac{1}{1 + 81 x ^{2}} d x

(2 x sin (y) cos (y)) y^{^{'}} = 4 x^{2} + sin^{2} (y)

\frac{d}{d x} (tan (e^{6 x}) + e^{t a n (6 x)})

\frac{\partial}{\partial y} (e^{x} sin (2 y))

What is the integral of (2x)/((2-x)^6) ?
The integral of (2x)/((2-x)^6) is -2(1/(4(2-x)^4)-2/(5(2-x)^5))+C