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

The integral of 1/(6x^2+x+1) is 2/(sqrt(23))arctan((12x+1)/(sqrt(23)))+C

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

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

\frac{2}{23} arctan (\frac{12 x + 1}{23}) + C

Solution steps

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

Complete the square

6 x^{2} + x + 1 : 6 (x + \frac{1}{12})^{2} + \frac{23}{24}

= \int \frac{1}{6 ( x + \frac{1}{12} ) ^{2} + \frac{23}{24}} d x

Apply u-substitution

: \int \frac{24}{144 u ^{2} + 23} d u

= \int \frac{24}{144 u ^{2} + 23} d u

Take the constant out:

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

= 24 \cdot \int \frac{1}{144 u ^{2} + 23} d u

Apply Integral Substitution

= 24 \cdot \int \frac{1}{12 23 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 24 \cdot \frac{1}{12 23} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 24 \cdot \frac{1}{12 23} arctan (v)

Substitute back

= 24 \cdot \frac{1}{12 23} arctan (\frac{12}{23} (x + \frac{1}{12}))

Simplify

24 \cdot \frac{1}{12 23} arctan (\frac{12}{23} (x + \frac{1}{12})) : \frac{2}{23} arctan (\frac{12 x + 1}{23})

= \frac{2}{23} arctan (\frac{12 x + 1}{23})

Add a constant to the solution

= \frac{2}{23} arctan (\frac{12 x + 1}{23}) + C

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What is the integral of 1/(6x^2+x+1) ?
The integral of 1/(6x^2+x+1) is 2/(sqrt(23))arctan((12x+1)/(sqrt(23)))+C