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

The integral of (x^2+2x-1)/(x(2x-1)(x+2)) is 1/2 ln|x|+1/10 ln|2x-1|-1/10 ln|x+2|+C

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

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

\frac{1}{2} ln ∣ x ∣ + \frac{1}{10} ln ∣ 2 x - 1 ∣ - \frac{1}{10} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 2 x - 1}{x ( 2 x - 1 ) ( x + 2 )} : \frac{1}{2 x} + \frac{1}{5 ( 2 x - 1 )} - \frac{1}{10 ( x + 2 )}

= \int \frac{1}{2 x} + \frac{1}{5 ( 2 x - 1 )} - \frac{1}{10 ( 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{1}{5 ( 2 x - 1 )} d x - \int \frac{1}{10 ( x + 2 )} d x

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

\int \frac{1}{5 ( 2 x - 1 )} d x = \frac{1}{10} ln ∣ 2 x - 1 ∣

\int \frac{1}{10 ( x + 2 )} d x = \frac{1}{10} ln ∣ x + 2 ∣

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{10} ln ∣ 2 x - 1 ∣ - \frac{1}{10} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{10} ln ∣ 2 x - 1 ∣ - \frac{1}{10} ln ∣ x + 2 ∣ + C

\int \frac{sin ( x )}{1 - cos ^{2} ( x )} d x

- 3 x^{2} - 10 x, 3

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

\int_{0}^{1} \frac{6}{x ^{2} + 2 x - 8} d x

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

What is the integral of (x^2+2x-1)/(x(2x-1)(x+2)) ?
The integral of (x^2+2x-1)/(x(2x-1)(x+2)) is 1/2 ln|x|+1/10 ln|2x-1|-1/10 ln|x+2|+C