What is the integral of x^3sqrt(x^2+15) ?

The integral of x^3sqrt(x^2+15) is 1/5 (x^2+15)^{5/2}-5(x^2+15)^{3/2}+C

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integral of x^3sqrt(x^2+15)

\int x^{3} x^{2} + 15 d x

\frac{1}{5} (x^{2} + 15)^{\frac{5}{2}} - 5 (x^{2} + 15)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 15 d x

Apply u-substitution

: \int (u^{2} - 15) u^{2} d u

= \int (u^{2} - 15) u^{2} d u

Expand

(u^{2} - 15) u^{2} : u^{4} - 15 u^{2}

= \int u^{4} - 15 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 15 u^{2} d u

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

\int 15 u^{2} d u = 5 u^{3}

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

Substitute back

u = x^{2} + 15

= \frac{( x ^{2} + 15 ) ^{5}}{5} - 5 (x^{2} + 15)^{3}

Simplify

\frac{( x ^{2} + 15 ) ^{5}}{5} - 5 (x^{2} + 15)^{3} : \frac{1}{5} (x^{2} + 15)^{\frac{5}{2}} - 5 (x^{2} + 15)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 15)^{\frac{5}{2}} - 5 (x^{2} + 15)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 15)^{\frac{5}{2}} - 5 (x^{2} + 15)^{\frac{3}{2}} + C

n = 1 \sum \infty \frac{1}{2 ^{n} ( n + 1 )}

x \to 9 lim (\frac{4 x}{x - 9})

y^{^{'}} \cdot (x - 1) + y \cdot (x - 2) = 0

\frac{1}{( s + 2 ) ( s + 3 )}

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

What is the integral of x^3sqrt(x^2+15) ?
The integral of x^3sqrt(x^2+15) is 1/5 (x^2+15)^{5/2}-5(x^2+15)^{3/2}+C