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

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

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integral of x/(x^2-2)

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

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

Solution steps

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

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

Add a constant to the solution

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

5 sin (5 t) sin (100 t)

\frac{6 t + 1}{6 t - 1}

x \to 0 lim (\frac{2 x ^{2} + 3 x}{x})

\frac{6 s + 6}{2 s ^{2} + 1}

y^{^{'}} e^{9 x} = 0

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