What is the integral of (x^2+2x-1)/(2x^3+3x^2-2x) ?

The integral of (x^2+2x-1)/(2x^3+3x^2-2x) 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)/(2x^3+3x^2-2x)

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

Take the partial fraction of

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

\frac{d}{d x} (x^{2} - 5 x + 4)

\frac{d y}{d x} = y^{2} - 9, y (5) = 3

\frac{\partial}{\partial x} (u)

y^{^{''}} - 2 y^{^{'}} + 26 y = 0

\int x^{8} 4 x^{3} + 4 d x

What is the integral of (x^2+2x-1)/(2x^3+3x^2-2x) ?
The integral of (x^2+2x-1)/(2x^3+3x^2-2x) is 1/2 ln|x|+1/10 ln|2x-1|-1/10 ln|x+2|+C