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

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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

= 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

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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