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

The integral of (x^3)/(x^2+4) is 1/2 (x^2+4-4ln|x^2+4|)+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int 1 - \frac{4}{u} 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

= \frac{1}{2} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{2} (u - 4 ln ∣ u ∣)

Substitute back

u = x^{2} + 4

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

Add a constant to the solution

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

5 sin (3 t - \frac{π}{4})

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

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

\frac{d}{d x} (25 x^{3} - 6 x^{2} + 7 x - 88)

What is the integral of (x^3)/(x^2+4) ?
The integral of (x^3)/(x^2+4) is 1/2 (x^2+4-4ln|x^2+4|)+C