What is the integral of (x^5)/(sqrt(1-x^2)) ?

The integral of (x^5)/(sqrt(1-x^2)) is -sqrt(1-x^2)+2/3 (1-x^2)^{3/2}-1/5 (1-x^2)^{5/2}+C

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integral of (x^5)/(sqrt(1-x^2))

\int \frac{x ^{5}}{1 - x ^{2}} d x

- 1 - x^{2} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + C

Solution steps

\int \frac{x ^{5}}{1 - x ^{2}} d x

Apply Trigonometric Substitution

: \int sin^{5} (u) d u

= \int sin^{5} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - 1 + 2 v^{2} - v^{4} d v

= \int - 1 + 2 v^{2} - v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= - v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5}

Substitute back

= - cos (arcsin (x)) + \frac{2 cos ^{3} ( arcsin ( x ) )}{3} - \frac{cos ^{5} ( arcsin ( x ) )}{5}

Simplify

- cos (arcsin (x)) + \frac{2 cos ^{3} ( arcsin ( x ) )}{3} - \frac{cos ^{5} ( arcsin ( x ) )}{5} : - 1 - x^{2} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}}

= - 1 - x^{2} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}}

Add a constant to the solution

= - 1 - x^{2} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}} - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + C

4^{x + 1}

x \to 1 lim (2 x^{3} - 4 x^{2} + x + 7)

\frac{\partial}{\partial x} (cos (2 y))

\int 39 - 10 x - x^{2} d x

\frac{d}{d x} (18 x - x^{2})

What is the integral of (x^5)/(sqrt(1-x^2)) ?
The integral of (x^5)/(sqrt(1-x^2)) is -sqrt(1-x^2)+2/3 (1-x^2)^{3/2}-1/5 (1-x^2)^{5/2}+C