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

The integral of (-9x^2)/((3-3x^3)^2) is

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

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

- 9 (- \frac{1}{81 ( x - 1 )} - \frac{2}{81 3} arctan (\frac{2 x + 1}{3}) - \frac{8}{27} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{x ^{2}}{( 3 - 3 x ^{3} ) ^{2}} : \frac{1}{81 ( x - 1 ) ^{2}} - \frac{1}{81 ( x ^{2} + x + 1 )} - \frac{x}{27 ( x ^{2} + x + 1 ) ^{2}}

= - 9 \cdot \int \frac{1}{81 ( x - 1 ) ^{2}} - \frac{1}{81 ( x ^{2} + x + 1 )} - \frac{x}{27 ( x ^{2} + x + 1 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 9 (\int \frac{1}{81 ( x - 1 ) ^{2}} d x - \int \frac{1}{81 ( x ^{2} + x + 1 )} d x - \int \frac{x}{27 ( x ^{2} + x + 1 ) ^{2}} d x)

\int \frac{1}{81 ( x - 1 ) ^{2}} d x = - \frac{1}{81 ( x - 1 )}

\int \frac{1}{81 ( x ^{2} + x + 1 )} d x = \frac{2}{81 3} arctan (\frac{2 x + 1}{3})

\int \frac{x}{27 ( x ^{2} + x + 1 ) ^{2}} d x = \frac{8}{27} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))

= - 9 (- \frac{1}{81 ( x - 1 )} - \frac{2}{81 3} arctan (\frac{2 x + 1}{3}) - \frac{8}{27} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))))

Add a constant to the solution

= - 9 (- \frac{1}{81 ( x - 1 )} - \frac{2}{81 3} arctan (\frac{2 x + 1}{3}) - \frac{8}{27} (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))) + C

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