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

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

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

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

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

Solution steps

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

\frac{\partial}{\partial y} (x y z^{2} tan (y t))

y = 2 - 6 x, at x = - 1

\frac{s}{s ^{2} - 3}

\int_{0}^{3} \frac{3}{2} x x + 1 d x

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