What is the integral of (6x^2)/((1-4x^3)^4) ?

The integral of (6x^2)/((1-4x^3)^4) is 1/(6(-4x^3+1)^3)+C

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integral of (6x^2)/((1-4x^3)^4)

\int \frac{6 x ^{2}}{( 1 - 4 x ^{3} ) ^{4}} d x

\frac{1}{6 ( - 4 x ^{3} + 1 ) ^{3}} + C

Solution steps

\int \frac{6 x ^{2}}{( 1 - 4 x ^{3} ) ^{4}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2}}{( 1 - 4 x ^{3} ) ^{4}} d x

Apply u-substitution

: \int \frac{u}{2 x ( 1 - 4 u ^{\frac{3}{2}} ) ^{4}} d u

= 6 \cdot \int \frac{1}{3 ( 1 - 4 u ) ^{4}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{( 1 - 4 u ) ^{4}} d u

Apply u-substitution

: \int - \frac{1}{4 v ^{4}} d v

= 6 \cdot \frac{1}{3} \cdot \int - \frac{1}{4 v ^{4}} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} (- \frac{1}{4} \cdot \int \frac{1}{v ^{4}} d v)

Apply the Power Rule

: - \frac{1}{3 v ^{3}}

= 6 \cdot \frac{1}{3} (- \frac{1}{4} (- \frac{1}{3 v ^{3}}))

Substitute back

= 6 \cdot \frac{1}{3} (- \frac{1}{4} (- \frac{1}{3 ( 1 - 4 x ^{3} ) ^{3}}))

Simplify

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

= \frac{1}{6 ( - 4 x ^{3} + 1 ) ^{3}}

Add a constant to the solution

= \frac{1}{6 ( - 4 x ^{3} + 1 ) ^{3}} + C

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What is the integral of (6x^2)/((1-4x^3)^4) ?
The integral of (6x^2)/((1-4x^3)^4) is 1/(6(-4x^3+1)^3)+C