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

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

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

\int \frac{x}{( x ^{2} - 4 )} d x

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 4

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

Add a constant to the solution

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

sin (8 x)

\int_{0}^{5} \frac{1}{x ^{\frac{13}{14}}} d x

y^{^{''}} + 2 y^{^{'}} - 2 y = 0

\frac{d}{d x} (ln (x x^{2} - 3))

x (30 - 2 x)^{2}

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