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

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

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

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

- ln ∣ 4 x - 2 ∣ + ln ∣ 4 x - 4 ∣ + C

Solution steps

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

Factor

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

= \int \frac{1}{- ( - 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

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

Complete the square

- 2 x^{2} + 3 x - 1 : - 2 (x - \frac{3}{4})^{2} + \frac{1}{8}

= - \int \frac{1}{- 2 ( x - \frac{3}{4} ) ^{2} + \frac{1}{8}} d x

Apply u-substitution

: \int \frac{8}{- 16 u ^{2} + 1} d u

= - \int \frac{8}{- 16 u ^{2} + 1} d u

Take the constant out:

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

= - 8 \cdot \int \frac{1}{- 16 u ^{2} + 1} d u

Apply Integral Substitution

= - 8 \cdot \int \frac{1}{4 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= - 8 \cdot \frac{1}{4} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - 8 \cdot \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - 8 \cdot \frac{1}{4} (\frac{ln 4 ( x - \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x - \frac{3}{4} ) - 1}{2})

Simplify

- 8 \cdot \frac{1}{4} (\frac{ln 4 ( x - \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x - \frac{3}{4} ) - 1}{2}) : - ln ∣ 4 x - 2 ∣ + ln ∣ 4 x - 4 ∣

= - ln ∣ 4 x - 2 ∣ + ln ∣ 4 x - 4 ∣

Add a constant to the solution

= - ln ∣ 4 x - 2 ∣ + ln ∣ 4 x - 4 ∣ + C

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