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

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

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

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

- \frac{1}{6 ( x ^{3} - 2 ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{3}} d v

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

Apply the Power Rule

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

= \frac{1}{3} (- \frac{1}{2 v ^{2}})

Substitute back

= \frac{1}{3} (- \frac{1}{2 ( x ^{3} - 2 ) ^{2}})

Simplify

\frac{1}{3} (- \frac{1}{2 ( x ^{3} - 2 ) ^{2}}) : - \frac{1}{6 ( x ^{3} - 2 ) ^{2}}

= - \frac{1}{6 ( x ^{3} - 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( x ^{3} - 2 ) ^{2}} + C

\frac{d}{d x} (\frac{8 x ^{2} + 4 x + 8}{x})

4 x^{3}, at x = 1

f (x) = 5 - 3 x, at x = \frac{1}{3}

tan (x)

f (x) = \frac{5}{r ^{3}}

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