What is the integral of 4/((x-6)^3) ?

The integral of 4/((x-6)^3) is -2/((x-6)^2)+C

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integral of 4/((x-6)^3)

\int \frac{4}{( x - 6 ) ^{3}} d x

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

Solution steps

\int \frac{4}{( x - 6 ) ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( x - 6 ) ^{3}} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = x - 6

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

Simplify

4 (- \frac{1}{2 ( x - 6 ) ^{2}}) : - \frac{2}{( x - 6 ) ^{2}}

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

Add a constant to the solution

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

100 x + \frac{d y}{d x} = x y

t \to 10 lim (\frac{t ^{2} + 3 t - 130}{t ^{2} - 100})

\frac{d}{d x} (\frac{e ^{- x}}{1 + x ^{2}})

\int \frac{sec ( x )}{( sec ( x ) - tan ( x ) )} d x

y^{^{'}} + 2 x = 0

What is the integral of 4/((x-6)^3) ?
The integral of 4/((x-6)^3) is -2/((x-6)^2)+C