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

The integral of (x-3)/(x^3+3x^2+2x) is -3/2 ln|x|+4ln|x+1|-5/2 ln|x+2|+C

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

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

- \frac{3}{2} ln ∣ x ∣ + 4 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x - 3}{x ^{3} + 3 x ^{2} + 2 x} : - \frac{3}{2 x} + \frac{4}{x + 1} - \frac{5}{2 ( x + 2 )}

= \int - \frac{3}{2 x} + \frac{4}{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{3}{2 x} d x + \int \frac{4}{x + 1} d x - \int \frac{5}{2 ( x + 2 )} d x

\int \frac{3}{2 x} d x = \frac{3}{2} ln ∣ x ∣

\int \frac{4}{x + 1} d x = 4 ln ∣ x + 1 ∣

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

= - \frac{3}{2} ln ∣ x ∣ + 4 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ x ∣ + 4 ln ∣ x + 1 ∣ - \frac{5}{2} ln ∣ x + 2 ∣ + C

\int e^{- \frac{2 r}{a}} d r

\frac{d}{d x} (\frac{1}{1 - 4 x})

\frac{d}{d x} (\frac{x ^{2}}{3 x - 4})

\int 2 x^{5} d x

x \to \infty lim (x_{1})

What is the integral of (x-3)/(x^3+3x^2+2x) ?
The integral of (x-3)/(x^3+3x^2+2x) is -3/2 ln|x|+4ln|x+1|-5/2 ln|x+2|+C