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

The integral of x/(x^2-3) is 1/2 ln|x^2-3|+C

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integral of x/(x^2-3)

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

\frac{1}{2} ln x^{2} - 3 + C

Solution steps

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

= \frac{1}{2} ln x^{2} - 3

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 3 + C

\frac{1}{x ( x ^{2} + 2 x + x )}

f (x) = 3 ln (x)

\frac{d}{d x} (\frac{1}{6 x + 5})

\frac{\partial}{\partial y} (7 ln (2 x^{5} - 3 y))

\frac{1}{10 x ^{2}}, x, \frac{x}{10}

What is the integral of x/(x^2-3) ?
The integral of x/(x^2-3) is 1/2 ln|x^2-3|+C