What is the integral of (4x^2)/((1-8x^3)^4) ?

The integral of (4x^2)/((1-8x^3)^4) is 1/(18(-8x^3+1)^3)+C

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integral of (4x^2)/((1-8x^3)^4)

\int \frac{4 x ^{2}}{( 1 - 8 x ^{3} ) ^{4}} d x

\frac{1}{18 ( - 8 x ^{3} + 1 ) ^{3}} + C

Solution steps

\int \frac{4 x ^{2}}{( 1 - 8 x ^{3} ) ^{4}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{( 1 - 8 x ^{3} ) ^{4}} d x

Apply u-substitution

: \int \frac{u}{2 x ( 1 - 8 u ^{\frac{3}{2}} ) ^{4}} d u

= 4 \cdot \int \frac{1}{3 ( 1 - 8 u ) ^{4}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{( 1 - 8 u ) ^{4}} d u

Apply u-substitution

: \int - \frac{1}{8 v ^{4}} d v

= 4 \cdot \frac{1}{3} \cdot \int - \frac{1}{8 v ^{4}} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} (- \frac{1}{8} \cdot \int \frac{1}{v ^{4}} d v)

Apply the Power Rule

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

= 4 \cdot \frac{1}{3} (- \frac{1}{8} (- \frac{1}{3 v ^{3}}))

Substitute back

= 4 \cdot \frac{1}{3} (- \frac{1}{8} (- \frac{1}{3 ( 1 - 8 x ^{3} ) ^{3}}))

Simplify

4 \cdot \frac{1}{3} (- \frac{1}{8} (- \frac{1}{3 ( 1 - 8 x ^{3} ) ^{3}})) : \frac{1}{18 ( - 8 x ^{3} + 1 ) ^{3}}

= \frac{1}{18 ( - 8 x ^{3} + 1 ) ^{3}}

Add a constant to the solution

= \frac{1}{18 ( - 8 x ^{3} + 1 ) ^{3}} + C

y^{^{'}} - \frac{y}{x} = 1

\int \frac{x}{a - x} d x

(x - 2 y)^{3} = 2 y^{2} - 3

\frac{d y}{d x}, x + y = 9

ln (e^{x})

What is the integral of (4x^2)/((1-8x^3)^4) ?
The integral of (4x^2)/((1-8x^3)^4) is 1/(18(-8x^3+1)^3)+C