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

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

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integral of x((1-x^2)^{3/2}-x^3)

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{x ( 5 - x )} d x

\int \frac{ln ( x + 9 )}{x + 9} d x

\int_{- 1}^{9} (9 - x^{2}) - (- 8 x) d x

y^{^{'}} = - 3 y^{2}

\frac{d}{d x} (\frac{1}{12} (x + 3)^{3} + (x + 3)^{- 1})

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