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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

y = - (2 x - 2)^{\frac{1}{3}}, (- 3, 2)

\frac{1}{2} \frac{d y}{d x} = y + 1 cos (x)

y = e^{2 x^{2}}

\int (x + x e^{x}) d x

x \to 0 lim ((x^{2} + 1)^{\frac{1}{x}})

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