What is the integral of x^3sqrt(x^2-81) ?

The integral of x^3sqrt(x^2-81) is 1/5 (x^2-81)^{5/2}+27(x^2-81)^{3/2}+C

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integral of x^3sqrt(x^2-81)

\int x^{3} x^{2} - 81 d x

\frac{1}{5} (x^{2} - 81)^{\frac{5}{2}} + 27 (x^{2} - 81)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} - 81 d x

Apply u-substitution

: \int (u^{2} + 81) u^{2} d u

= \int (u^{2} + 81) u^{2} d u

Expand

(u^{2} + 81) u^{2} : u^{4} + 81 u^{2}

= \int u^{4} + 81 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 81 u^{2} d u

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

\int 81 u^{2} d u = 27 u^{3}

= \frac{u ^{5}}{5} + 27 u^{3}

Substitute back

u = x^{2} - 81

= \frac{( x ^{2} - 81 ) ^{5}}{5} + 27 (x^{2} - 81)^{3}

Simplify

\frac{( x ^{2} - 81 ) ^{5}}{5} + 27 (x^{2} - 81)^{3} : \frac{1}{5} (x^{2} - 81)^{\frac{5}{2}} + 27 (x^{2} - 81)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} - 81)^{\frac{5}{2}} + 27 (x^{2} - 81)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 81)^{\frac{5}{2}} + 27 (x^{2} - 81)^{\frac{3}{2}} + C

\int \frac{cot ^{3} ( x )}{5} d x

\frac{d}{d y} ((\frac{y}{8})^{4} + \frac{1}{4 y ^{2}})

x^{2} y^{^{''}} - 2 y = 0

x \to 1 lim (cos (π) x)

n = 1 \sum \infty 4 (\frac{2}{3})^{n - 1}

What is the integral of x^3sqrt(x^2-81) ?
The integral of x^3sqrt(x^2-81) is 1/5 (x^2-81)^{5/2}+27(x^2-81)^{3/2}+C