What is the integral of (5x^3)/((x^4-1)^3) ?

The integral of (5x^3)/((x^4-1)^3) is -5/(8(x^4-1)^2)+C

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integral of (5x^3)/((x^4-1)^3)

\int \frac{5 x ^{3}}{( x ^{4} - 1 ) ^{3}} d x

- \frac{5}{8 ( x ^{4} - 1 ) ^{2}} + C

Solution steps

\int \frac{5 x ^{3}}{( x ^{4} - 1 ) ^{3}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{3}}{( x ^{4} - 1 ) ^{3}} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{4} - 1

= 5 \cdot \frac{1}{4} (- \frac{1}{2 ( x ^{4} - 1 ) ^{2}})

Simplify

5 \cdot \frac{1}{4} (- \frac{1}{2 ( x ^{4} - 1 ) ^{2}}) : - \frac{5}{8 ( x ^{4} - 1 ) ^{2}}

= - \frac{5}{8 ( x ^{4} - 1 ) ^{2}}

Add a constant to the solution

= - \frac{5}{8 ( x ^{4} - 1 ) ^{2}} + C

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What is the integral of (5x^3)/((x^4-1)^3) ?
The integral of (5x^3)/((x^4-1)^3) is -5/(8(x^4-1)^2)+C