What is the integral of (8x^2+16x)(4x^3+12x^2)^{-3} ?

The integral of (8x^2+16x)(4x^3+12x^2)^{-3} is -1/(3(4x^3+12x^2)^2)+C

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integral of (8x^2+16x)(4x^3+12x^2)^{-3}

\int (8 x^{2} + 16 x) (4 x^{3} + 12 x^{2})^{- 3} d x

- \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}} + C

Solution steps

\int (8 x^{2} + 16 x) (4 x^{3} + 12 x^{2})^{- 3} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 4 x^{3} + 12 x^{2}

= \frac{2}{3} (- \frac{1}{2 ( 4 x ^{3} + 12 x ^{2} ) ^{2}})

Simplify

\frac{2}{3} (- \frac{1}{2 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}) : - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}

= - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}}

Add a constant to the solution

= - \frac{1}{3 ( 4 x ^{3} + 12 x ^{2} ) ^{2}} + C

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What is the integral of (8x^2+16x)(4x^3+12x^2)^{-3} ?
The integral of (8x^2+16x)(4x^3+12x^2)^{-3} is -1/(3(4x^3+12x^2)^2)+C