What is the integral of (x^3)/(sqrt((4-x^2)^3)) ?

The integral of (x^3)/(sqrt((4-x^2)^3)) is (x^4)/(4sqrt(4-x^2))-1/4 (4-x^2)^{3/2}+3sqrt(4-x^2)+C

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integral of (x^3)/(sqrt((4-x^2)^3))

\int \frac{x ^{3}}{( 4 - x ^{2} ) ^{3}} d x

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

Solution steps

\int \frac{x ^{3}}{( 4 - x ^{2} ) ^{3}} d x

Apply Integration By Parts

: \frac{x ^{4}}{4 4 - x ^{2}} - \int \frac{3 x ^{3}}{4 4 - x ^{2}} d x

= \frac{x ^{4}}{4 4 - x ^{2}} - \int \frac{3 x ^{3}}{4 4 - x ^{2}} d x

\int \frac{3 x ^{3}}{4 4 - x ^{2}} d x = \frac{1}{4} (4 - x^{2})^{\frac{3}{2}} - 3 4 - x^{2}

= \frac{x ^{4}}{4 4 - x ^{2}} - (\frac{1}{4} (4 - x^{2})^{\frac{3}{2}} - 3 4 - x^{2})

Simplify

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

Add a constant to the solution

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

y^{^{'}} = (x + y - 1)^{2} - 1

\frac{\partial}{\partial x} (\frac{t}{x - 1})

\int 2 x - 2 y d x

\frac{1}{1 - z ^{2}}, 0

\frac{\partial}{\partial x} (\frac{1}{2} ln (x))

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