What is the integral of (-10x^2)/((x^3+2)^{-3)} ?

The integral of (-10x^2)/((x^3+2)^{-3)} is -5/6 (x^3+2)^4+C

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integral of (-10x^2)/((x^3+2)^{-3)}

\int \frac{- 10 x ^{2}}{( x ^{3} + 2 ) ^{- 3}} d x

- \frac{5}{6} (x^{3} + 2)^{4} + C

Solution steps

\int \frac{- 10 x ^{2}}{( x ^{3} + 2 ) ^{- 3}} d x

Take the constant out:

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

= - 10 \cdot \int \frac{x ^{2}}{( x ^{3} + 2 ) ^{- 3}} d x

Simplify

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

= - 10 \cdot \int x^{2} (x^{3} + 2)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{3} d u

= - 10 \cdot \int \frac{u ^{3}}{3} d u

Take the constant out:

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

= - 10 \cdot \frac{1}{3} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - 10 \cdot \frac{1}{3} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{3} + 2

= - 10 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 2 ) ^{4}}{4}

Simplify

- 10 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 2 ) ^{4}}{4} : - \frac{5}{6} (x^{3} + 2)^{4}

= - \frac{5}{6} (x^{3} + 2)^{4}

Add a constant to the solution

= - \frac{5}{6} (x^{3} + 2)^{4} + C

\frac{d}{d x} (\frac{250}{( 1 + 5 x ) ^{3}})

y^{^{'}} = 6 x^{2} - x^{2} y

\int e^{3 t} \cdot t d t

\frac{d}{d x} (\frac{1}{( 1 + 2 x )})

\frac{d}{d x} ((- 16 x^{2} + 36)^{2})

What is the integral of (-10x^2)/((x^3+2)^{-3)} ?
The integral of (-10x^2)/((x^3+2)^{-3)} is -5/6 (x^3+2)^4+C