What is the integral of (x^2)/((6-x^3)^2) ?

The integral of (x^2)/((6-x^3)^2) is 1/(3(6-x^3))+C

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integral of (x^2)/((6-x^3)^2)

\int \frac{x ^{2}}{( 6 - x ^{3} ) ^{2}} d x

\frac{1}{3 ( 6 - x ^{3} )} + C

Solution steps

\int \frac{x ^{2}}{( 6 - x ^{3} ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 6 - x^{3}

= - \frac{1}{3} (- \frac{1}{6 - x ^{3}})

Simplify

- \frac{1}{3} (- \frac{1}{6 - x ^{3}}) : \frac{1}{3 ( 6 - x ^{3} )}

= \frac{1}{3 ( 6 - x ^{3} )}

Add a constant to the solution

= \frac{1}{3 ( 6 - x ^{3} )} + C

n = 0 \sum \infty \frac{3}{( n + 1 )}

\frac{x ^{3}}{2} + \frac{1}{6 x}

\frac{d}{d x} (3 cos (x sin (x)))

n = 1 \sum \infty \frac{n}{n ^{2} + 9}

f (x) = 2 x^{2} - 5 x, at x = 3

What is the integral of (x^2)/((6-x^3)^2) ?
The integral of (x^2)/((6-x^3)^2) is 1/(3(6-x^3))+C