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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int u^{2} - 1 d u

= - \int u^{2} - 1 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

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

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

\int 1 d u = u

= - (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{a x^{5}})

x \to 0 lim (2 x (1 - ln (x)))

\frac{d}{d x} (arcsin (x^{9}))

f (x) = 6 x^{2} - 4 x

\frac{\partial}{\partial x} (x (5 z^{4} - 4 y^{3}))

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