What is the integral of (x^8)/(sqrt(1+x^3)) ?

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

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integral of (x^8)/(sqrt(1+x^3))

\int \frac{x ^{8}}{1 + x ^{3}} d x

\frac{1}{3} (\frac{2}{5} (1 + x^{3})^{\frac{5}{2}} - \frac{4}{3} (1 + x^{3})^{\frac{3}{2}} + 2 1 + x^{3}) + C

Solution steps

\int \frac{x ^{8}}{1 + x ^{3}} d x

Apply u-substitution

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

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

Expand

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

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

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

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}} + 2 u)

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (\frac{2}{5} (1 + x^{3})^{\frac{5}{2}} - \frac{4}{3} (1 + x^{3})^{\frac{3}{2}} + 2 1 + x^{3})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (1 + x^{3})^{\frac{5}{2}} - \frac{4}{3} (1 + x^{3})^{\frac{3}{2}} + 2 1 + x^{3}) + C

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