What is the integral of (8x+12)/(x*(x^2+4)) ?

The integral of (8x+12)/(x*(x^2+4)) is 4arctan(x/2)+12(1/4 ln|x|-1/8 ln|x^2+4|)+C

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integral of (8x+12)/(x*(x^2+4))

\int \frac{8 x + 12}{x \cdot ( x ^{2} + 4 )} d x

4 arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

Solution steps

\int \frac{8 x + 12}{x ( x ^{2} + 4 )} d x

Expand

\frac{8 x + 12}{x ( x ^{2} + 4 )} : \frac{8 x}{x ( x ^{2} + 4 )} + \frac{12}{x ( x ^{2} + 4 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8 x}{x ( x ^{2} + 4 )} d x + \int \frac{12}{x ( x ^{2} + 4 )} d x

\int \frac{8 x}{x ( x ^{2} + 4 )} d x = 4 arctan (\frac{x}{2})

\int \frac{12}{x ( x ^{2} + 4 )} d x = 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= 4 arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Add a constant to the solution

= 4 arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

\frac{\partial}{\partial x} (x 1 + y^{3})

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

3 x^{2} - 1, at x = - 2

y = \frac{x ^{2} + 2}{x + 8}, (3, 1)

\int 2 e^{3 x - 5} d x

What is the integral of (8x+12)/(x*(x^2+4)) ?
The integral of (8x+12)/(x*(x^2+4)) is 4arctan(x/2)+12(1/4 ln|x|-1/8 ln|x^2+4|)+C