What is the integral of (x^3)/((2+3x^4)^3) ?

The integral of (x^3)/((2+3x^4)^3) is -1/(24(2+3x^4)^2)+C

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integral of (x^3)/((2+3x^4)^3)

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

- \frac{1}{24 ( 2 + 3 x ^{4} ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{12 u ^{3}} d u

= \int \frac{1}{12 u ^{3}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= \frac{1}{12} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 + 3 x^{4}

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

Simplify

\frac{1}{12} (- \frac{1}{2 ( 2 + 3 x ^{4} ) ^{2}}) : - \frac{1}{24 ( 2 + 3 x ^{4} ) ^{2}}

= - \frac{1}{24 ( 2 + 3 x ^{4} ) ^{2}}

Add a constant to the solution

= - \frac{1}{24 ( 2 + 3 x ^{4} ) ^{2}} + C

n = 1 \sum \infty \frac{7}{n}

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

(8 + x) y^{^{'}} = 5 y

\frac{10}{( s ( s ^{2} + 2 s + 20 ) )}

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

What is the integral of (x^3)/((2+3x^4)^3) ?
The integral of (x^3)/((2+3x^4)^3) is -1/(24(2+3x^4)^2)+C