What is the integral of (2x-1)/(x(x^2+3x+2)) ?

The integral of (2x-1)/(x(x^2+3x+2)) is -1/2 ln|x|+3ln|x+1|-5/2 ln|x+2|+C

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integral of (2x-1)/(x(x^2+3x+2))

\int \frac{2 x - 1}{x ( x ^{2} + 3 x + 2 )} d x

- \frac{1}{2} ln ∣ x ∣ + 3 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{2 x - 1}{x ( x ^{2} + 3 x + 2 )} d x

Take the partial fraction of

\frac{2 x - 1}{x ( x ^{2} + 3 x + 2 )} : - \frac{1}{2 x} + \frac{3}{x + 1} - \frac{5}{2 ( x + 2 )}

= \int - \frac{1}{2 x} + \frac{3}{x + 1} - \frac{5}{2 ( x + 2 )} d x

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{1}{2 x} d x + \int \frac{3}{x + 1} d x - \int \frac{5}{2 ( x + 2 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{5}{2 ( x + 2 )} d x = \frac{5}{2} ln ∣ x + 2 ∣

= - \frac{1}{2} ln ∣ x ∣ + 3 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + 3 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣ + C

2 \frac{d y}{d t} + y = 0

\frac{d}{d x} (\frac{x - 3}{x - 4})

\frac{d}{d x} (e^{x} - π)

\int \frac{x ^{2}}{16 + x ^{2}} d x

\frac{1}{x}, x, \frac{1}{4} x

What is the integral of (2x-1)/(x(x^2+3x+2)) ?
The integral of (2x-1)/(x(x^2+3x+2)) is -1/2 ln|x|+3ln|x+1|-5/2 ln|x+2|+C