What is the integral of x/((3-x)^7) ?

The integral of x/((3-x)^7) is -1/(5(3-x)^5)+1/(2(3-x)^6)+C

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integral of x/((3-x)^7)

\int \frac{x}{( 3 - x ) ^{7}} d x

- \frac{1}{5 ( 3 - x ) ^{5}} + \frac{1}{2 ( 3 - x ) ^{6}} + C

Solution steps

\int \frac{x}{( 3 - x ) ^{7}} d x

Apply u-substitution

: \int - \frac{- u + 3}{u ^{7}} d u

= \int - \frac{- u + 3}{u ^{7}} d u

Take the constant out:

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

= - \int \frac{- u + 3}{u ^{7}} d u

Expand

\frac{- u + 3}{u ^{7}} : - \frac{1}{u ^{6}} + \frac{3}{u ^{7}}

= - \int - \frac{1}{u ^{6}} + \frac{3}{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

= - (- \int \frac{1}{u ^{6}} d u + \int \frac{3}{u ^{7}} d u)

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

\int \frac{3}{u ^{7}} d u = - \frac{1}{2 u ^{6}}

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

Substitute back

u = 3 - x

= - (- (- \frac{1}{5 ( 3 - x ) ^{5}}) - \frac{1}{2 ( 3 - x ) ^{6}})

Simplify

= - \frac{1}{5 ( 3 - x ) ^{5}} + \frac{1}{2 ( 3 - x ) ^{6}}

Add a constant to the solution

= - \frac{1}{5 ( 3 - x ) ^{5}} + \frac{1}{2 ( 3 - x ) ^{6}} + C

\int \frac{1}{( x ^{4} + 4 )} d x

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What is the integral of x/((3-x)^7) ?
The integral of x/((3-x)^7) is -1/(5(3-x)^5)+1/(2(3-x)^6)+C