What is the integral of x^3sqrt(x^2+4) ?

The integral of x^3sqrt(x^2+4) is 1/5 (x^2+4)^{5/2}-4/3 (x^2+4)^{3/2}+C

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integral of x^3sqrt(x^2+4)

\int x^{3} x^{2} + 4 d x

\frac{1}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 4 d x

Apply u-substitution

: \int (u^{2} - 4) u^{2} d u

= \int (u^{2} - 4) u^{2} d u

Expand

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

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

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

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{4 u ^{3}}{3}

Substitute back

u = x^{2} + 4

= \frac{( x ^{2} + 4 ) ^{5}}{5} - \frac{4 ( x ^{2} + 4 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 4 ) ^{5}}{5} - \frac{4 ( x ^{2} + 4 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{2} + 4)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{2} + 4)^{\frac{3}{2}} + C

\frac{d y}{d x} - 2 x y = 15 x^{2} e^{x^{2}}

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

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 5

\frac{d}{d x} (7 e^{x} - (\frac{10}{x ^{4}}))

f (x) = \frac{2 - \frac{1}{x}}{x - 3}

What is the integral of x^3sqrt(x^2+4) ?
The integral of x^3sqrt(x^2+4) is 1/5 (x^2+4)^{5/2}-4/3 (x^2+4)^{3/2}+C