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

\int \frac{3 x + 1}{5 x ^{2} + 1} d x

x \to - 1 lim (- 2)

f (x) = 2 x^{2} + 3 x - 9

\int \frac{ln ( x ^{8} )}{x} d x

\frac{1}{3 x - 1}

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