What is the integral of (1-x^3)^2x ?

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

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integral of (1-x^3)^2x

\int (1 - x^{3})^{2} x d x

\frac{x ^{2}}{2} - \frac{2 x ^{5}}{5} + \frac{x ^{8}}{8} + C

Solution steps

\int (1 - x^{3})^{2} x d x

Expand

(1 - x^{3})^{2} x : x - 2 x^{4} + x^{7}

= \int x - 2 x^{4} + x^{7} d x

Apply the Sum Rule:

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

= \int x d x - \int 2 x^{4} d x + \int x^{7} d x

\int x d x = \frac{x ^{2}}{2}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int x^{7} d x = \frac{x ^{8}}{8}

= \frac{x ^{2}}{2} - \frac{2 x ^{5}}{5} + \frac{x ^{8}}{8}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{2 x ^{5}}{5} + \frac{x ^{8}}{8} + C

\int x e^{- 5 x^{2}} d x

\int (2 x + 4) (x^{2} + 4 x + 10)^{4} d x

\int x^{5} x^{3} + 3 d x

\int_{0}^{1} 2 x^{4} e^{- x} d x

y = (x^{2} - 3)^{4}, at x = 2

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