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

The integral of 1/(3x^2-x+1) is 2/(sqrt(11))arctan((6x-1)/(sqrt(11)))+C

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

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

\frac{2}{11} arctan (\frac{6 x - 1}{11}) + C

Solution steps

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

Complete the square

3 x^{2} - x + 1 : 3 (x - \frac{1}{6})^{2} + \frac{11}{12}

= \int \frac{1}{3 ( x - \frac{1}{6} ) ^{2} + \frac{11}{12}} d x

Apply u-substitution

: \int \frac{12}{36 u ^{2} + 11} d u

= \int \frac{12}{36 u ^{2} + 11} d u

Take the constant out:

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

= 12 \cdot \int \frac{1}{36 u ^{2} + 11} d u

Apply Integral Substitution

= 12 \cdot \int \frac{1}{6 11 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 12 \cdot \frac{1}{6 11} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 12 \cdot \frac{1}{6 11} arctan (v)

Substitute back

= 12 \cdot \frac{1}{6 11} arctan (\frac{6}{11} (x - \frac{1}{6}))

Simplify

12 \cdot \frac{1}{6 11} arctan (\frac{6}{11} (x - \frac{1}{6})) : \frac{2}{11} arctan (\frac{6 x - 1}{11})

= \frac{2}{11} arctan (\frac{6 x - 1}{11})

Add a constant to the solution

= \frac{2}{11} arctan (\frac{6 x - 1}{11}) + C

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What is the integral of 1/(3x^2-x+1) ?
The integral of 1/(3x^2-x+1) is 2/(sqrt(11))arctan((6x-1)/(sqrt(11)))+C