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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (π (6 x - y)))

arctan (\frac{1 + t}{1 - t})

\frac{d}{d x} (- 2 (arctan (3 x) - arctan (2 x)) + x)

\int \frac{3}{60 - t} d t

\frac{8}{x ^{2} + 4}

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