What is the integral of x/((2x+6)^7) ?

The integral of x/((2x+6)^7) is (-2x-1)/(1280(x+3)^6)+C

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integral of x/((2x+6)^7)

\int \frac{x}{( 2 x + 6 ) ^{7}} d x

\frac{- 2 x - 1}{1280 ( x + 3 ) ^{6}} + C

Solution steps

\int \frac{x}{( 2 x + 6 ) ^{7}} d x

Apply u-substitution

: \int \frac{u - 6}{4 u ^{7}} d u

= \int \frac{u - 6}{4 u ^{7}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 6}{u ^{7}} d u

Expand

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{6}} - \frac{6}{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

= \frac{1}{4} (\int \frac{1}{u ^{6}} d u - \int \frac{6}{u ^{7}} d u)

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

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

= \frac{1}{4} (- \frac{1}{5 u ^{5}} - (- \frac{1}{u ^{6}}))

Substitute back

u = 2 x + 6

= \frac{1}{4} (- \frac{1}{5 ( 2 x + 6 ) ^{5}} - (- \frac{1}{( 2 x + 6 ) ^{6}}))

Simplify

\frac{1}{4} (- \frac{1}{5 ( 2 x + 6 ) ^{5}} - (- \frac{1}{( 2 x + 6 ) ^{6}})) : \frac{- 2 x - 1}{1280 ( x + 3 ) ^{6}}

= \frac{- 2 x - 1}{1280 ( x + 3 ) ^{6}}

Add a constant to the solution

= \frac{- 2 x - 1}{1280 ( x + 3 ) ^{6}} + C

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What is the integral of x/((2x+6)^7) ?
The integral of x/((2x+6)^7) is (-2x-1)/(1280(x+3)^6)+C