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

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

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

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

x^{2} - 4 + 4 ln x^{2} - 4 + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{3}}{x ^{2} - 4} d x

Apply u-substitution

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

= 2 \cdot \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

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

Expand

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

= 2 \cdot \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

= 2 \cdot \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 ∣

= 2 \cdot \frac{1}{2} (u + 4 ln ∣ u ∣)

Substitute back

u = x^{2} - 4

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

Simplify

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

= x^{2} - 4 + 4 ln x^{2} - 4

Add a constant to the solution

= x^{2} - 4 + 4 ln x^{2} - 4 + C

\frac{d}{d x} (∣ x - 3 ∣ - ∣ 2 x - 2 ∣ - 1 x e [- 2, 2])

\frac{d}{d x} (1 + x ln (x y - 5))

y^{^{'''}} + y^{^{'}} = 0

x \to 2 lim (3 x^{2} + 2 x - 1)

\int \frac{1}{x ^{2} + a} d x

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