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

The integral of (x^5)/((x^3+1)^5) is (-4x^3-1)/(36(x+1)^4(x^2-x+1)^4)+C

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integral of (x^5)/((x^3+1)^5)

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

\frac{- 4 x ^{3} - 1}{36 ( x + 1 ) ^{4} ( x ^{2} - x + 1 ) ^{4}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{3 u ^{5}} d u

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

Take the constant out:

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

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

Expand

\frac{u - 1}{u ^{5}} : \frac{1}{u ^{4}} - \frac{1}{u ^{5}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{1}{3} (- \frac{1}{3 u ^{3}} - (- \frac{1}{4 u ^{4}}))

Substitute back

u = x^{3} + 1

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

Simplify

\frac{1}{3} (- \frac{1}{3 ( x ^{3} + 1 ) ^{3}} - (- \frac{1}{4 ( x ^{3} + 1 ) ^{4}})) : \frac{- 4 x ^{3} - 1}{36 ( x + 1 ) ^{4} ( x ^{2} - x + 1 ) ^{4}}

= \frac{- 4 x ^{3} - 1}{36 ( x + 1 ) ^{4} ( x ^{2} - x + 1 ) ^{4}}

Add a constant to the solution

= \frac{- 4 x ^{3} - 1}{36 ( x + 1 ) ^{4} ( x ^{2} - x + 1 ) ^{4}} + C

\frac{1}{cos ( x )}

\int x^{2} x^{3} - 15 d x

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What is the integral of (x^5)/((x^3+1)^5) ?
The integral of (x^5)/((x^3+1)^5) is (-4x^3-1)/(36(x+1)^4(x^2-x+1)^4)+C