What is the integral of (2r)/((1-r)^7) ?

The integral of (2r)/((1-r)^7) is -2(1/(5(1-r)^5)-1/(6(1-r)^6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2r)/((1-r)^7)

\int \frac{2 r}{( 1 - r ) ^{7}} d r

- 2 (\frac{1}{5 ( 1 - r ) ^{5}} - \frac{1}{6 ( 1 - r ) ^{6}}) + C

Solution steps

\int \frac{2 r}{( 1 - r ) ^{7}} d r

Take the constant out:

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

= 2 \cdot \int \frac{r}{( 1 - r ) ^{7}} d r

Apply u-substitution

: \int - \frac{- u + 1}{u ^{7}} d u

= 2 \cdot \int - \frac{- u + 1}{u ^{7}} d u

Take the constant out:

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

= 2 (- \int \frac{- u + 1}{u ^{7}} d u)

Expand

\frac{- u + 1}{u ^{7}} : - \frac{1}{u ^{6}} + \frac{1}{u ^{7}}

= 2 (- \int - \frac{1}{u ^{6}} + \frac{1}{u ^{7}} 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 ^{6}} d u + \int \frac{1}{u ^{7}} d u))

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

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

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

Substitute back

u = 1 - r

= 2 (- (- (- \frac{1}{5 ( 1 - r ) ^{5}}) - \frac{1}{6 ( 1 - r ) ^{6}}))

Simplify

= - 2 (\frac{1}{5 ( 1 - r ) ^{5}} - \frac{1}{6 ( 1 - r ) ^{6}})

Add a constant to the solution

= - 2 (\frac{1}{5 ( 1 - r ) ^{5}} - \frac{1}{6 ( 1 - r ) ^{6}}) + C

y = x^{2}, y = 2 - x

\int - \frac{2}{1 - 9 x ^{2}} d x

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

\int_{- 1}^{2} - 2 e^{1 - 3 x} d x

n = 1 \sum \infty cos (nπ)

What is the integral of (2r)/((1-r)^7) ?
The integral of (2r)/((1-r)^7) is -2(1/(5(1-r)^5)-1/(6(1-r)^6))+C