What is the integral of (x^3)/((3-2x)^5) ?

The integral of (x^3)/((3-2x)^5) is (72x^2-144x-4(-2x+3)^3+81)/(64(-2x+3)^4)+C

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integral of (x^3)/((3-2x)^5)

\int \frac{x ^{3}}{( 3 - 2 x ) ^{5}} d x

\frac{72 x ^{2} - 144 x - 4 ( - 2 x + 3 ) ^{3} + 81}{64 ( - 2 x + 3 ) ^{4}} + C

Solution steps

\int \frac{x ^{3}}{( 3 - 2 x ) ^{5}} d x

Apply u-substitution

: \int \frac{( u - 3 ) ^{3}}{16 u ^{5}} d u

= \int \frac{( u - 3 ) ^{3}}{16 u ^{5}} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{( u - 3 ) ^{3}}{u ^{5}} d u

Expand

\frac{( u - 3 ) ^{3}}{u ^{5}} : \frac{1}{u ^{2}} - \frac{9}{u ^{3}} + \frac{27}{u ^{4}} - \frac{27}{u ^{5}}

= \frac{1}{16} \cdot \int \frac{1}{u ^{2}} - \frac{9}{u ^{3}} + \frac{27}{u ^{4}} - \frac{27}{u ^{5}} 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}{16} (\int \frac{1}{u ^{2}} d u - \int \frac{9}{u ^{3}} d u + \int \frac{27}{u ^{4}} d u - \int \frac{27}{u ^{5}} d u)

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

\int \frac{9}{u ^{3}} d u = - \frac{9}{2 u ^{2}}

\int \frac{27}{u ^{4}} d u = - \frac{9}{u ^{3}}

\int \frac{27}{u ^{5}} d u = - \frac{27}{4 u ^{4}}

= \frac{1}{16} (- \frac{1}{u} - (- \frac{9}{2 u ^{2}}) - \frac{9}{u ^{3}} - (- \frac{27}{4 u ^{4}}))

Substitute back

u = 3 - 2 x

= \frac{1}{16} (- \frac{1}{3 - 2 x} - (- \frac{9}{2 ( 3 - 2 x ) ^{2}}) - \frac{9}{( 3 - 2 x ) ^{3}} - (- \frac{27}{4 ( 3 - 2 x ) ^{4}}))

Simplify

\frac{1}{16} (- \frac{1}{3 - 2 x} - (- \frac{9}{2 ( 3 - 2 x ) ^{2}}) - \frac{9}{( 3 - 2 x ) ^{3}} - (- \frac{27}{4 ( 3 - 2 x ) ^{4}})) : \frac{72 x ^{2} - 144 x - 4 ( - 2 x + 3 ) ^{3} + 81}{64 ( - 2 x + 3 ) ^{4}}

= \frac{72 x ^{2} - 144 x - 4 ( - 2 x + 3 ) ^{3} + 81}{64 ( - 2 x + 3 ) ^{4}}

Add a constant to the solution

= \frac{72 x ^{2} - 144 x - 4 ( - 2 x + 3 ) ^{3} + 81}{64 ( - 2 x + 3 ) ^{4}} + C

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What is the integral of (x^3)/((3-2x)^5) ?
The integral of (x^3)/((3-2x)^5) is (72x^2-144x-4(-2x+3)^3+81)/(64(-2x+3)^4)+C