What is the integral of x^3sqrt(9+x^2) ?

The integral of x^3sqrt(9+x^2) is 1/5 (9+x^2)^{5/2}-3(9+x^2)^{3/2}+C

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integral of x^3sqrt(9+x^2)

\int x^{3} 9 + x^{2} d x

\frac{1}{5} (9 + x^{2})^{\frac{5}{2}} - 3 (9 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 9 + x^{2} d x

Apply u-substitution

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

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

Expand

(u^{2} - 9) u^{2} : u^{4} - 9 u^{2}

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

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

\int 9 u^{2} d u = 3 u^{3}

= \frac{u ^{5}}{5} - 3 u^{3}

Substitute back

u = 9 + x^{2}

= \frac{( 9 + x ^{2} ) ^{5}}{5} - 3 (9 + x^{2})^{3}

Simplify

\frac{( 9 + x ^{2} ) ^{5}}{5} - 3 (9 + x^{2})^{3} : \frac{1}{5} (9 + x^{2})^{\frac{5}{2}} - 3 (9 + x^{2})^{\frac{3}{2}}

= \frac{1}{5} (9 + x^{2})^{\frac{5}{2}} - 3 (9 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (9 + x^{2})^{\frac{5}{2}} - 3 (9 + x^{2})^{\frac{3}{2}} + C

\int 5 sin (8 x) cos (5 x) d x

\frac{1}{5 x + 7}

\frac{\partial}{\partial x} (arcsin (11 x yz))

\frac{d}{d x} (\frac{5 x}{4 x ^{2} + 1})

What is the integral of x^3sqrt(9+x^2) ?
The integral of x^3sqrt(9+x^2) is 1/5 (9+x^2)^{5/2}-3(9+x^2)^{3/2}+C