What is the integral of (4x^3)/(sqrt(9-x^2)) ?

The integral of (4x^3)/(sqrt(9-x^2)) is 4(1/3 (9-x^2)^{3/2}-9sqrt(9-x^2))+C

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integral of (4x^3)/(sqrt(9-x^2))

\int \frac{4 x ^{3}}{9 - x ^{2}} d x

4 (\frac{1}{3} (9 - x^{2})^{\frac{3}{2}} - 9 9 - x^{2}) + C

Solution steps

\int \frac{4 x ^{3}}{9 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x ^{3}}{9 - x ^{2}} d x

Apply u-substitution

: \int u^{2} - 9 d u

= 4 \cdot \int u^{2} - 9 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

= 4 (\int u^{2} d u - \int 9 d u)

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

\int 9 d u = 9 u

= 4 (\frac{u ^{3}}{3} - 9 u)

Substitute back

u = 9 - x^{2}

= 4 (\frac{( 9 - x ^{2} ) ^{3}}{3} - 9 9 - x^{2})

Simplify

4 (\frac{( 9 - x ^{2} ) ^{3}}{3} - 9 9 - x^{2}) : 4 (\frac{1}{3} (9 - x^{2})^{\frac{3}{2}} - 9 9 - x^{2})

= 4 (\frac{1}{3} (9 - x^{2})^{\frac{3}{2}} - 9 9 - x^{2})

Add a constant to the solution

= 4 (\frac{1}{3} (9 - x^{2})^{\frac{3}{2}} - 9 9 - x^{2}) + C

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What is the integral of (4x^3)/(sqrt(9-x^2)) ?
The integral of (4x^3)/(sqrt(9-x^2)) is 4(1/3 (9-x^2)^{3/2}-9sqrt(9-x^2))+C