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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - (- u^{2} + 3) u^{2} d u

= \int - (- u^{2} + 3) u^{2} d u

Take the constant out:

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

= - \int (- u^{2} + 3) u^{2} d u

Expand

(- u^{2} + 3) u^{2} : - u^{4} + 3 u^{2}

= - \int - u^{4} + 3 u^{2} 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^{4} d u + \int 3 u^{2} d u)

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

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

= - (- \frac{u ^{5}}{5} + u^{3})

Substitute back

u = 3 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{6}}{1 - x ^{5}})

e^{x} y^{^{'}} = 2 x

\int \frac{4}{x ^{2} + 9} d x

6 x y

\int x x - 30 d x

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