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

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

- x^{4} + 2 x^{3} - 5 x^{2} + 0.4

\int_{1}^{4} 3 x d x

\frac{s + 5}{s ( s + 10 ) ^{2}}

\frac{x y ^{^{'}}}{e ^{y} + 1} = 2

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