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

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

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

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

- \frac{1}{3 ( x ^{3} - 3 )} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{3 ( u - 3 ) ^{2}} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{3} (- \frac{1}{v})

Substitute back

= \frac{1}{3} (- \frac{1}{x ^{3} - 3})

Simplify

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

= - \frac{1}{3 ( x ^{3} - 3 )}

Add a constant to the solution

= - \frac{1}{3 ( x ^{3} - 3 )} + C

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