What is the integral of x^4(x^5-2)^3 ?

The integral of x^4(x^5-2)^3 is 1/20 (x^{20}-8x^{15}+24x^{10}-32x^5)+C

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integral of x^4(x^5-2)^3

\int x^{4} (x^{5} - 2)^{3} d x

\frac{1}{20} (x^{20} - 8 x^{15} + 24 x^{10} - 32 x^{5}) + C

Solution steps

\int x^{4} (x^{5} - 2)^{3} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{4}} ( u ^{\frac{5}{4}} - 2 ) ^{3}}{4} d u

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

Expand

u^{\frac{1}{4}} (u^{\frac{5}{4}} - 2)^{3} : u^{4} - 6 u^{\frac{11}{4}} + 12 u^{\frac{3}{2}} - 8 u^{\frac{1}{4}}

= \frac{1}{4} \cdot \int u^{4} - 6 u^{\frac{11}{4}} + 12 u^{\frac{3}{2}} - 8 u^{\frac{1}{4}} 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}{4} (\int u^{4} d u - \int 6 u^{\frac{11}{4}} d u + \int 12 u^{\frac{3}{2}} d u - \int 8 u^{\frac{1}{4}} d u)

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

\int 6 u^{\frac{11}{4}} d u = \frac{8}{5} u^{\frac{15}{4}}

\int 12 u^{\frac{3}{2}} d u = \frac{24}{5} u^{\frac{5}{2}}

\int 8 u^{\frac{1}{4}} d u = \frac{32}{5} u^{\frac{5}{4}}

= \frac{1}{4} (\frac{u ^{5}}{5} - \frac{8}{5} u^{\frac{15}{4}} + \frac{24}{5} u^{\frac{5}{2}} - \frac{32}{5} u^{\frac{5}{4}})

Substitute back

u = x^{4}

= \frac{1}{4} (\frac{( x ^{4} ) ^{5}}{5} - \frac{8}{5} (x^{4})^{\frac{15}{4}} + \frac{24}{5} (x^{4})^{\frac{5}{2}} - \frac{32}{5} (x^{4})^{\frac{5}{4}})

Simplify

\frac{1}{4} (\frac{( x ^{4} ) ^{5}}{5} - \frac{8}{5} (x^{4})^{\frac{15}{4}} + \frac{24}{5} (x^{4})^{\frac{5}{2}} - \frac{32}{5} (x^{4})^{\frac{5}{4}}) : \frac{1}{20} (x^{20} - 8 x^{15} + 24 x^{10} - 32 x^{5})

= \frac{1}{20} (x^{20} - 8 x^{15} + 24 x^{10} - 32 x^{5})

Add a constant to the solution

= \frac{1}{20} (x^{20} - 8 x^{15} + 24 x^{10} - 32 x^{5}) + C

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What is the integral of x^4(x^5-2)^3 ?
The integral of x^4(x^5-2)^3 is 1/20 (x^{20}-8x^{15}+24x^{10}-32x^5)+C