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

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

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integral of 3/(x^2-4x+1)

\int \frac{3}{x ^{2} - 4 x + 1} d x

- 3 (\frac{1}{2} ln \frac{x - 2}{3} + 1 - \frac{1}{2} ln \frac{x - 2}{3} - 1) + C

Solution steps

\int \frac{3}{x ^{2} - 4 x + 1} 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} - 4 x + 1} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 3} d u

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

Apply Integral Substitution

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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