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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 4 cos (x))

x \to 1 lim (4 x - 5)

\frac{\partial}{\partial z} (x y e^{yz})

x \to 3 lim (∣ x - 2 ∣)

\int (2 + 2 x) d x

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