What is the integral of 8/((4-x)^3) ?

The integral of 8/((4-x)^3) is 4/((4-x)^2)+C

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integral of 8/((4-x)^3)

\int \frac{8}{( 4 - x ) ^{3}} d x

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

Solution steps

\int \frac{8}{( 4 - x ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{( 4 - x ) ^{3}} d x

Apply u-substitution

: \int - \frac{1}{u ^{3}} d u

= 8 \cdot \int - \frac{1}{u ^{3}} d u

Take the constant out:

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

= 8 (- \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 8 (- (- \frac{1}{2 u ^{2}}))

Substitute back

u = 4 - x

= 8 (- (- \frac{1}{2 ( 4 - x ) ^{2}}))

Simplify

8 (- (- \frac{1}{2 ( 4 - x ) ^{2}})) : \frac{4}{( 4 - x ) ^{2}}

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

Add a constant to the solution

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

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What is the integral of 8/((4-x)^3) ?
The integral of 8/((4-x)^3) is 4/((4-x)^2)+C