What is the integral of (x^4)/(sqrt(1-x^2)) ?

The integral of (x^4)/(sqrt(1-x^2)) is -1/4 x^3sqrt(1-x^2)+3/8 (arcsin(x)-1/2 sin(2arcsin(x)))+C

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integral of (x^4)/(sqrt(1-x^2))

\int \frac{x ^{4}}{1 - x ^{2}} d x

- \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + C

Solution steps

\int \frac{x ^{4}}{1 - x ^{2}} d x

Apply Trigonometric Substitution

: \int sin^{4} (u) d u

= \int sin^{4} (u) d u

Apply Integral Reduction

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u

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

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (x)

= - \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

Simplify

- \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) : - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

= - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

Add a constant to the solution

= - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + C

(x^{3} - 12 x) (3 x^{2} + 2 x)

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

(3 x e^{x})^{^{'}}

\int x \frac{d}{1} (y) d x

\frac{\partial}{\partial x} (x \cdot ln (x + 2 y) + \frac{x}{y})

What is the integral of (x^4)/(sqrt(1-x^2)) ?
The integral of (x^4)/(sqrt(1-x^2)) is -1/4 x^3sqrt(1-x^2)+3/8 (arcsin(x)-1/2 sin(2arcsin(x)))+C