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

The integral of 3/(x^2-3x+2) is -6(1/2 ln|2x-2|-1/2 ln|2x-4|)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 1} d u

= 3 \cdot \int \frac{4}{4 u ^{2} - 1} d u

Take the constant out:

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

= 3 \cdot 4 \cdot \int \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

= 3 \cdot 4 (- \frac{1}{2} \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}

= 3 \cdot 4 (- \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

3 \cdot 4 (- \frac{1}{2} (\frac{ln 2 ( x - \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( x - \frac{3}{2} ) - 1}{2})) : - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

= - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

Add a constant to the solution

= - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣) + C

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What is the integral of 3/(x^2-3x+2) ?
The integral of 3/(x^2-3x+2) is -6(1/2 ln|2x-2|-1/2 ln|2x-4|)+C