What is the integral of x^3sqrt(16+x^2) ?

The integral of x^3sqrt(16+x^2) is 1/5 (16+x^2)^{5/2}-16/3 (16+x^2)^{3/2}+C

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integral of x^3sqrt(16+x^2)

\int x^{3} 16 + x^{2} d x

\frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 16 + x^{2} d x

Apply u-substitution

: \int (u^{2} - 16) u^{2} d u

= \int (u^{2} - 16) u^{2} d u

Expand

(u^{2} - 16) u^{2} : u^{4} - 16 u^{2}

= \int u^{4} - 16 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 16 u^{2} d u

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

\int 16 u^{2} d u = \frac{16 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{16 u ^{3}}{3}

Substitute back

u = 16 + x^{2}

= \frac{( 16 + x ^{2} ) ^{5}}{5} - \frac{16 ( 16 + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 16 + x ^{2} ) ^{5}}{5} - \frac{16 ( 16 + x ^{2} ) ^{3}}{3} : \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}}

= \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}} + C

\frac{x}{1 - ln ( x - 8 )}

4 x (x^{2} - 9)

x \to 0 lim (\frac{8 x}{cot ( x )})

3 x^{2} + x, at x = 1

(- 6, 4), (- 5, - 10)

What is the integral of x^3sqrt(16+x^2) ?
The integral of x^3sqrt(16+x^2) is 1/5 (16+x^2)^{5/2}-16/3 (16+x^2)^{3/2}+C