What is the integral of (-15x^3)/((3x^4+2)^{-2)} ?

The integral of (-15x^3)/((3x^4+2)^{-2)} is -60((3x^{12}}{16}+\frac{3x^8)/8+(x^4)/4)+C

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integral of (-15x^3)/((3x^4+2)^{-2)}

\int \frac{- 15 x ^{3}}{( 3 x ^{4} + 2 ) ^{- 2}} d x

- 60 (\frac{3 x ^{12}}{16} + \frac{3 x ^{8}}{8} + \frac{x ^{4}}{4}) + C

Solution steps

\int \frac{- 15 x ^{3}}{( 3 x ^{4} + 2 ) ^{- 2}} d x

Take the constant out:

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

= - 15 \cdot \int \frac{x ^{3}}{( 3 x ^{4} + 2 ) ^{- 2}} d x

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

= - 15 \cdot \int \frac{x ^{3}}{( 2 ( \frac{3 x ^{4}}{2} + 1 ) ) ^{- 2}} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{3 x ^{4}}{2} = (\frac{4 3 x}{4 2})^{4}

= - 15 \cdot \int \frac{x ^{3}}{( 2 ( ( \frac{4 3 x}{4 2} ) ^{4} + 1 ) ) ^{- 2}} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

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

= - 15 \cdot \int \frac{x ^{3}}{2 ^{- 2} ( ( \frac{4 3 x}{4 2} ) ^{4} + 1 ) ^{- 2}} d x

Take the constant out:

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

= - 15 \cdot \frac{1}{2 ^{- 2}} \cdot \int \frac{x ^{3}}{( ( \frac{4 3 x}{4 2} ) ^{4} + 1 ) ^{- 2}} d x

\frac{1}{2 ^{- 2}} = 4

= - 15 \cdot 4 \cdot \int \frac{x ^{3}}{( ( \frac{4 3 x}{4 2} ) ^{4} + 1 ) ^{- 2}} d x

Expand

\frac{x ^{3}}{( ( \frac{4 3 x}{4 2} ) ^{4} + 1 ) ^{- 2}} : \frac{9 x ^{11}}{4} + 3 x^{7} + x^{3}

= - 15 \cdot 4 \cdot \int \frac{9 x ^{11}}{4} + 3 x^{7} + x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 15 \cdot 4 (\int \frac{9 x ^{11}}{4} d x + \int 3 x^{7} d x + \int x^{3} d x)

\int \frac{9 x ^{11}}{4} d x = \frac{3 x ^{12}}{16}

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

\int x^{3} d x = \frac{x ^{4}}{4}

= - 15 \cdot 4 (\frac{3 x ^{12}}{16} + \frac{3 x ^{8}}{8} + \frac{x ^{4}}{4})

Simplify

= - 60 (\frac{3 x ^{12}}{16} + \frac{3 x ^{8}}{8} + \frac{x ^{4}}{4})

Add a constant to the solution

= - 60 (\frac{3 x ^{12}}{16} + \frac{3 x ^{8}}{8} + \frac{x ^{4}}{4}) + C

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What is the integral of (-15x^3)/((3x^4+2)^{-2)} ?
The integral of (-15x^3)/((3x^4+2)^{-2)} is -60((3x^{12}}{16}+\frac{3x^8)/8+(x^4)/4)+C