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

The integral of ((x+3))/(x^3+x^2-2x) is -3/2 ln|x|+4/3 ln|x-1|+1/6 ln|x+2|+C

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

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

- \frac{3}{2} ln ∣ x ∣ + \frac{4}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 3}{x ^{3} + x ^{2} - 2 x} : - \frac{3}{2 x} + \frac{4}{3 ( x - 1 )} + \frac{1}{6 ( x + 2 )}

= \int - \frac{3}{2 x} + \frac{4}{3 ( x - 1 )} + \frac{1}{6 ( 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}{3 ( x - 1 )} d x + \int \frac{1}{6 ( x + 2 )} d x

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

\int \frac{4}{3 ( x - 1 )} d x = \frac{4}{3} ln ∣ x - 1 ∣

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

= - \frac{3}{2} ln ∣ x ∣ + \frac{4}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ x ∣ + \frac{4}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣ + C

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in t e g r a l - cos^{2} (x)

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What is the integral of ((x+3))/(x^3+x^2-2x) ?
The integral of ((x+3))/(x^3+x^2-2x) is -3/2 ln|x|+4/3 ln|x-1|+1/6 ln|x+2|+C