What is the integral of (x^3)/((3x^4-1)^3) ?

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

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integral of (x^3)/((3x^4-1)^3)

\int \frac{x ^{3}}{( 3 x ^{4} - 1 ) ^{3}} d x

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

Solution steps

\int \frac{x ^{3}}{( 3 x ^{4} - 1 ) ^{3}} d x

Apply u-substitution

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

Substitute back

u = 3 x^{4} - 1

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

Simplify

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

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

Add a constant to the solution

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

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