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

The integral of x^3(2-x^2)^2 is (x^8)/8-2/3 x^6+x^4+C

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

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

\frac{x ^{8}}{8} - \frac{2}{3} x^{6} + x^{4} + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

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

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

= \frac{1}{3} (\frac{3}{8} u^{\frac{8}{3}} - 2 u^{2} + 3 u^{\frac{4}{3}})

Substitute back

u = x^{3}

= \frac{1}{3} (\frac{3}{8} (x^{3})^{\frac{8}{3}} - 2 (x^{3})^{2} + 3 (x^{3})^{\frac{4}{3}})

Simplify

\frac{1}{3} (\frac{3}{8} (x^{3})^{\frac{8}{3}} - 2 (x^{3})^{2} + 3 (x^{3})^{\frac{4}{3}}) : \frac{x ^{8}}{8} - \frac{2}{3} x^{6} + x^{4}

= \frac{x ^{8}}{8} - \frac{2}{3} x^{6} + x^{4}

Add a constant to the solution

= \frac{x ^{8}}{8} - \frac{2}{3} x^{6} + x^{4} + C

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What is the integral of x^3(2-x^2)^2 ?
The integral of x^3(2-x^2)^2 is (x^8)/8-2/3 x^6+x^4+C