What is the integral of (5x+12)/(x(x^2+4)) ?

The integral of (5x+12)/(x(x^2+4)) is 5/2 arctan(x/2)+12(1/4 ln|x|-1/8 ln|x^2+4|)+C

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integral of (5x+12)/(x(x^2+4))

\int \frac{5 x + 12}{x ( x ^{2} + 4 )} d x

\frac{5}{2} arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

Solution steps

\int \frac{5 x + 12}{x ( x ^{2} + 4 )} d x

Expand

\frac{5 x + 12}{x ( x ^{2} + 4 )} : \frac{5 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{5 x}{x ( x ^{2} + 4 )} d x + \int \frac{12}{x ( x ^{2} + 4 )} d x

\int \frac{5 x}{x ( x ^{2} + 4 )} d x = \frac{5}{2} 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)

= \frac{5}{2} arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Add a constant to the solution

= \frac{5}{2} arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) + C

\int (x + 1)^{- 8} x d x

ln (4 x + 5)

y^{^{''}} - 49 y = 0

x \to 8 lim (7 x - 4)

n = 0 \sum \infty 1.00 2^{n}

What is the integral of (5x+12)/(x(x^2+4)) ?
The integral of (5x+12)/(x(x^2+4)) is 5/2 arctan(x/2)+12(1/4 ln|x|-1/8 ln|x^2+4|)+C