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

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

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

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

- \frac{1}{8 ( x ^{4} - 4 x ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u ^{3}} d u

= \int \frac{1}{4 u ^{3}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{4} - 4 x

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

Simplify

\frac{1}{4} (- \frac{1}{2 ( x ^{4} - 4 x ) ^{2}}) : - \frac{1}{8 ( x ^{4} - 4 x ) ^{2}}

= - \frac{1}{8 ( x ^{4} - 4 x ) ^{2}}

Add a constant to the solution

= - \frac{1}{8 ( x ^{4} - 4 x ) ^{2}} + C

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