What is the integral of (x^3)/(sqrt(16-x^2)) ?

The integral of (x^3)/(sqrt(16-x^2)) is 1/3 (16-x^2)^{3/2}-16sqrt(16-x^2)+C

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integral of (x^3)/(sqrt(16-x^2))

\int \frac{x ^{3}}{16 - x ^{2}} d x

\frac{1}{3} (16 - x^{2})^{\frac{3}{2}} - 16 16 - x^{2} + C

Solution steps

\int \frac{x ^{3}}{16 - x ^{2}} d x

Apply u-substitution

: \int u^{2} - 16 d u

= \int u^{2} - 16 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 16 d u

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

\int 16 d u = 16 u

= \frac{u ^{3}}{3} - 16 u

Substitute back

u = 16 - x^{2}

= \frac{( 16 - x ^{2} ) ^{3}}{3} - 16 16 - x^{2}

Simplify

\frac{( 16 - x ^{2} ) ^{3}}{3} - 16 16 - x^{2} : \frac{1}{3} (16 - x^{2})^{\frac{3}{2}} - 16 16 - x^{2}

= \frac{1}{3} (16 - x^{2})^{\frac{3}{2}} - 16 16 - x^{2}

Add a constant to the solution

= \frac{1}{3} (16 - x^{2})^{\frac{3}{2}} - 16 16 - x^{2} + C

8 x

\int \frac{x ^{4} - 7 x ^{2} + 5}{x ^{4}} d x

\int_{0}^{3} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

y^{^{'}} = \frac{t ^{3} + 19 t ^{3} y}{t}

\frac{\partial}{\partial y} (- yz)

What is the integral of (x^3)/(sqrt(16-x^2)) ?
The integral of (x^3)/(sqrt(16-x^2)) is 1/3 (16-x^2)^{3/2}-16sqrt(16-x^2)+C