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

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

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

\int \frac{( x ^{3} )}{x ^{2} + 4} d x

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

Solution steps

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

Apply u-substitution

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

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

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

\int 4 d u = 4 u

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

Substitute back

u = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

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

4 (t - 2)

\int - 2 y^{3} d y

\frac{d}{d x} (\frac{1 - 6 ln ( x )}{x ^{7}})

\frac{d}{d x} ((sin (x))^{\frac{1}{x}})

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