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

\frac{cos ( 2 x )}{3}

y^{^{'}} - t y^{3} = 0, y (0) = 2

\int sin (t) tan (t) d t

14 x (x^{2} + 4)^{6}

5 x^{2} cos (x) cot (x)

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