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

\frac{\partial}{\partial x} (4 x y + x^{4} + y^{4})

\frac{d}{d x} (y - x)

x \to 1 lim (\frac{x ^{2}}{x ^{3} - x})

\frac{d}{d x} (cos (x) tan (x))

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

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