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

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

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

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

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

Solution steps

\int x^{3} x^{2} - 16 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 = x^{2} - 16

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (5 - ∣ x ∣)

\int tan (3 x se) c^{2} 3 x d x

x \to - 3 lim (x^{2} + x - x)

\int_{0}^{1} \frac{x}{x ^{3} + 1} d x

\frac{d}{d x} (\frac{x - 6}{x ^{2} - 2 x + 12})

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