What is the integral of (x^3)/(sqrt(25-x^2)) ?

The integral of (x^3)/(sqrt(25-x^2)) is 1/3 (25-x^2)^{3/2}-25sqrt(25-x^2)+C

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integral of (x^3)/(sqrt(25-x^2))

\int \frac{x ^{3}}{25 - x ^{2}} d x

\frac{1}{3} (25 - x^{2})^{\frac{3}{2}} - 25 25 - x^{2} + C

Solution steps

\int \frac{x ^{3}}{25 - x ^{2}} d x

Apply u-substitution

: \int u^{2} - 25 d u

= \int u^{2} - 25 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 25 d u

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

\int 25 d u = 25 u

= \frac{u ^{3}}{3} - 25 u

Substitute back

u = 25 - x^{2}

= \frac{( 25 - x ^{2} ) ^{3}}{3} - 25 25 - x^{2}

Simplify

\frac{( 25 - x ^{2} ) ^{3}}{3} - 25 25 - x^{2} : \frac{1}{3} (25 - x^{2})^{\frac{3}{2}} - 25 25 - x^{2}

= \frac{1}{3} (25 - x^{2})^{\frac{3}{2}} - 25 25 - x^{2}

Add a constant to the solution

= \frac{1}{3} (25 - x^{2})^{\frac{3}{2}} - 25 25 - x^{2} + C

\int_{0}^{2} 3 y 4 - y^{2} d y

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3

x \frac{d y}{d x} - 2 y = x^{2}

\frac{d}{d y} (- e^{- x y})

\int \frac{25 - x ^{2}}{x} d x

What is the integral of (x^3)/(sqrt(25-x^2)) ?
The integral of (x^3)/(sqrt(25-x^2)) is 1/3 (25-x^2)^{3/2}-25sqrt(25-x^2)+C