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

The integral of (x^5)/((1+x^3)^3) is (-2x^3-1)/(6(x+1)^2(x^2-x+1)^2)+C

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

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

\frac{- 2 x ^{3} - 1}{6 ( x + 1 ) ^{2} ( x ^{2} - x + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= \frac{1}{3} \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{3}} 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 ^{2}} d u - \int \frac{1}{u ^{3}} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= \frac{1}{3} (- \frac{1}{u} - (- \frac{1}{2 u ^{2}}))

Substitute back

u = 1 + x^{3}

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

Simplify

\frac{1}{3} (- \frac{1}{1 + x ^{3}} - (- \frac{1}{2 ( 1 + x ^{3} ) ^{2}})) : \frac{- 2 x ^{3} - 1}{6 ( x + 1 ) ^{2} ( x ^{2} - x + 1 ) ^{2}}

= \frac{- 2 x ^{3} - 1}{6 ( x + 1 ) ^{2} ( x ^{2} - x + 1 ) ^{2}}

Add a constant to the solution

= \frac{- 2 x ^{3} - 1}{6 ( x + 1 ) ^{2} ( x ^{2} - x + 1 ) ^{2}} + C

What is the integral of (x^5)/((1+x^3)^3) ?
The integral of (x^5)/((1+x^3)^3) is (-2x^3-1)/(6(x+1)^2(x^2-x+1)^2)+C