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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

\int \frac{1}{2 ( 2 x - 1 )} d x = \frac{1}{4} ln ∣ 2 x - 1 ∣

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

= 2 (\frac{1}{2} x - \frac{1}{4} ln ∣ 2 x - 1 ∣ + ln ∣ x - 1 ∣)

Simplify

2 (\frac{1}{2} x - \frac{1}{4} ln ∣ 2 x - 1 ∣ + ln ∣ x - 1 ∣) : x - \frac{1}{2} ln ∣ 2 x - 1 ∣ + 2 ln ∣ x - 1 ∣

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

Add a constant to the solution

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

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