What is the integral of (5x+3)/(x^3-2x^2-3x) ?

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

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integral of (5x+3)/(x^3-2x^2-3x)

\int \frac{5 x + 3}{x ^{3} - 2 x ^{2} - 3 x} d x

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

Solution steps

\int \frac{5 x + 3}{x ^{3} - 2 x ^{2} - 3 x} d x

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

y y^{^{'}} = x (y - 1)

6 x^{2}, at x = 10

22 cos (t)

\int x^{2} - 4 x + 2 d x

\int \frac{3}{x} d x

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