What is the integral of (3x)/(x^2-4) ?

The integral of (3x)/(x^2-4) is 3/2 ln|x^2-4|+C

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integral of (3x)/(x^2-4)

\int \frac{3 x}{x ^{2} - 4} d x

\frac{3}{2} ln x^{2} - 4 + C

Solution steps

\int \frac{3 x}{x ^{2} - 4} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} - 4} d x

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 4

= 3 \cdot \frac{1}{2} ln x^{2} - 4

Simplify

3 \cdot \frac{1}{2} ln x^{2} - 4 : \frac{3}{2} ln x^{2} - 4

= \frac{3}{2} ln x^{2} - 4

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 4 + C

\int (x - \frac{1}{x})^{2} d x

y^{^{'}} = (x - 5) e^{- 2 y}

\int_{1}^{2} \frac{v ^{4} + 5 v ^{7}}{v ^{3}} d v

- t^{2} δ (t - 2)

\int x^{5} x^{3} + 1 d x

What is the integral of (3x)/(x^2-4) ?
The integral of (3x)/(x^2-4) is 3/2 ln|x^2-4|+C