What is the integral of x^3sqrt(x^2-5) ?

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

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integral of x^3sqrt(x^2-5)

\int x^{3} x^{2} - 5 d x

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

Solution steps

\int x^{3} x^{2} - 5 d x

Apply u-substitution

: \int (u^{2} + 5) u^{2} d u

= \int (u^{2} + 5) u^{2} d u

Expand

(u^{2} + 5) u^{2} : u^{4} + 5 u^{2}

= \int u^{4} + 5 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 5 u^{2} d u

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

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

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

Substitute back

u = x^{2} - 5

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} - \frac{3}{x} y = x^{3}

k = 1 \sum \infty 0

y = x, x = 3, x = 6

x^{3} + 9 x^{2} + 35 x + 14

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

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