What is the integral of x^3sqrt(x^2+27) ?

The integral of x^3sqrt(x^2+27) is 1/5 (x^2+27)^{5/2}-9(x^2+27)^{3/2}+C

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integral of x^3sqrt(x^2+27)

\int x^{3} x^{2} + 27 d x

\frac{1}{5} (x^{2} + 27)^{\frac{5}{2}} - 9 (x^{2} + 27)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 27 d x

Apply u-substitution

: \int (u^{2} - 27) u^{2} d u

= \int (u^{2} - 27) u^{2} d u

Expand

(u^{2} - 27) u^{2} : u^{4} - 27 u^{2}

= \int u^{4} - 27 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 27 u^{2} d u

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

\int 27 u^{2} d u = 9 u^{3}

= \frac{u ^{5}}{5} - 9 u^{3}

Substitute back

u = x^{2} + 27

= \frac{( x ^{2} + 27 ) ^{5}}{5} - 9 (x^{2} + 27)^{3}

Simplify

\frac{( x ^{2} + 27 ) ^{5}}{5} - 9 (x^{2} + 27)^{3} : \frac{1}{5} (x^{2} + 27)^{\frac{5}{2}} - 9 (x^{2} + 27)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 27)^{\frac{5}{2}} - 9 (x^{2} + 27)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 27)^{\frac{5}{2}} - 9 (x^{2} + 27)^{\frac{3}{2}} + C

\frac{( a + h ) ^{4}}{h}

\frac{d}{d x} ((x^{4} - 7 x)^{2})

y^{^{'}} = y (x y^{3} - 1)

f (x) = 2^{80}

x \to 1 lim (1 + \frac{1}{x - 1})

What is the integral of x^3sqrt(x^2+27) ?
The integral of x^3sqrt(x^2+27) is 1/5 (x^2+27)^{5/2}-9(x^2+27)^{3/2}+C