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

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

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

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

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

Solution steps

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

Expand

(1 - x^{2} - x) x : x 1 - x^{2} - x^{2}

= \int x 1 - x^{2} - x^{2} 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} d x - \int x^{2} d x

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

Add a constant to the solution

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

\int \frac{x ^{4} + x - 16}{x ^{2} + 4} d x

ln (\frac{a ^{2} - z ^{2}}{a ^{2} + z ^{2}})

y = 3 (4 x^{2} + 4)

\frac{\partial}{\partial x} (z x + y)

\frac{\partial}{\partial x} ((x^{2} + y^{2}))

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