What is the integral of 1/(x^2-x+6) ?

The integral of 1/(x^2-x+6) is 2/(sqrt(23))arctan((2x-1)/(sqrt(23)))+C

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integral of 1/(x^2-x+6)

\int \frac{1}{x ^{2} - x + 6} d x

\frac{2}{23} arctan (\frac{2 x - 1}{23}) + C

Solution steps

\int \frac{1}{x ^{2} - x + 6} d x

Complete the square

x^{2} - x + 6 : (x - \frac{1}{2})^{2} + \frac{23}{4}

= \int \frac{1}{( x - \frac{1}{2} ) ^{2} + \frac{23}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 23} d u

= \int \frac{4}{4 u ^{2} + 23} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 23} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 23 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 23} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot \frac{1}{2 23} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2 23} arctan (\frac{2}{23} (x - \frac{1}{2}))

Simplify

4 \cdot \frac{1}{2 23} arctan (\frac{2}{23} (x - \frac{1}{2})) : \frac{2}{23} arctan (\frac{2 x - 1}{23})

= \frac{2}{23} arctan (\frac{2 x - 1}{23})

Add a constant to the solution

= \frac{2}{23} arctan (\frac{2 x - 1}{23}) + C

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What is the integral of 1/(x^2-x+6) ?
The integral of 1/(x^2-x+6) is 2/(sqrt(23))arctan((2x-1)/(sqrt(23)))+C