What is the integral of (8x)/(x^2-4) ?

The integral of (8x)/(x^2-4) is 4ln|x^2-4|+C

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integral of (8x)/(x^2-4)

\int \frac{8 x}{x ^{2} - 4} d x

4 ln x^{2} - 4 + C

Solution steps

\int \frac{8 x}{x ^{2} - 4} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{x ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 8 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 4

= 8 \cdot \frac{1}{2} ln x^{2} - 4

Simplify

8 \cdot \frac{1}{2} ln x^{2} - 4 : 4 ln x^{2} - 4

= 4 ln x^{2} - 4

Add a constant to the solution

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

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What is the integral of (8x)/(x^2-4) ?
The integral of (8x)/(x^2-4) is 4ln|x^2-4|+C