What is the integral of (4x)/(2x^2-6) ?

The integral of (4x)/(2x^2-6) is ln|2x^2-6|+C

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integral of (4x)/(2x^2-6)

\int \frac{4 x}{2 x ^{2} - 6} d x

ln 2 x^{2} - 6 + C

Solution steps

\int \frac{4 x}{2 x ^{2} - 6} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{2 x ^{2} - 6} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 4 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} - 6

= 4 \cdot \frac{1}{4} ln 2 x^{2} - 6

Simplify

4 \cdot \frac{1}{4} ln 2 x^{2} - 6 : ln 2 x^{2} - 6

= ln 2 x^{2} - 6

Add a constant to the solution

= ln 2 x^{2} - 6 + C

x \cdot \frac{d y}{d x} + y = \frac{2 y}{1 - y ^{2}}

y = x - 3, y = 2 x + 2, y = - 2 x + 2

\int (\frac{2}{x}) d x

\int_{1}^{\infty} \frac{7}{x ^{3}} d x

\frac{\partial}{\partial x} (\frac{1}{x - 2})

What is the integral of (4x)/(2x^2-6) ?
The integral of (4x)/(2x^2-6) is ln|2x^2-6|+C