What is the integral of 1/((1-2/9 x)^3) ?

The integral of 1/((1-2/9 x)^3) is (729)/(4(2x-9)^2)+C

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integral of 1/((1-2/9 x)^3)

\int \frac{1}{( 1 - \frac{2}{9} x ) ^{3}} d x

\frac{729}{4 ( 2 x - 9 ) ^{2}} + C

Solution steps

\int \frac{1}{( 1 - \frac{2}{9} x ) ^{3}} d x

Take the partial fraction of

\frac{1}{( 1 - \frac{2}{9} x ) ^{3}} : - \frac{729}{( 2 x - 9 ) ^{3}}

= \int - \frac{729}{( 2 x - 9 ) ^{3}} d x

Take the constant out:

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

= - 729 \cdot \int \frac{1}{( 2 x - 9 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= - 729 \cdot \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= - 729 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= - 729 \cdot \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 x - 9

= - 729 \cdot \frac{1}{2} (- \frac{1}{2 ( 2 x - 9 ) ^{2}})

Simplify

- 729 \cdot \frac{1}{2} (- \frac{1}{2 ( 2 x - 9 ) ^{2}}) : \frac{729}{4 ( 2 x - 9 ) ^{2}}

= \frac{729}{4 ( 2 x - 9 ) ^{2}}

Add a constant to the solution

= \frac{729}{4 ( 2 x - 9 ) ^{2}} + C

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What is the integral of 1/((1-2/9 x)^3) ?
The integral of 1/((1-2/9 x)^3) is (729)/(4(2x-9)^2)+C