What is the integral of x^3sqrt(x^2-6) ?

The integral of x^3sqrt(x^2-6) is 1/5 (x^2-6)^{5/2}+2(x^2-6)^{3/2}+C

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integral of x^3sqrt(x^2-6)

\int x^{3} x^{2} - 6 d x

\frac{1}{5} (x^{2} - 6)^{\frac{5}{2}} + 2 (x^{2} - 6)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} - 6 d x

Apply u-substitution

: \int (u^{2} + 6) u^{2} d u

= \int (u^{2} + 6) u^{2} d u

Expand

(u^{2} + 6) u^{2} : u^{4} + 6 u^{2}

= \int u^{4} + 6 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 6 u^{2} d u

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

\int 6 u^{2} d u = 2 u^{3}

= \frac{u ^{5}}{5} + 2 u^{3}

Substitute back

u = x^{2} - 6

= \frac{( x ^{2} - 6 ) ^{5}}{5} + 2 (x^{2} - 6)^{3}

Simplify

\frac{( x ^{2} - 6 ) ^{5}}{5} + 2 (x^{2} - 6)^{3} : \frac{1}{5} (x^{2} - 6)^{\frac{5}{2}} + 2 (x^{2} - 6)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} - 6)^{\frac{5}{2}} + 2 (x^{2} - 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 6)^{\frac{5}{2}} + 2 (x^{2} - 6)^{\frac{3}{2}} + C

\frac{d}{d t} (5^{t})

x \to 1 - lim (sech^{2} (x - 1))

\frac{\partial}{\partial y} (2 x y^{3} + 4)

\int 12 + 4 t^{2} d t

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 9

What is the integral of x^3sqrt(x^2-6) ?
The integral of x^3sqrt(x^2-6) is 1/5 (x^2-6)^{5/2}+2(x^2-6)^{3/2}+C