What is the integral of 1/(6x^2+3x+1) ?

The integral of 1/(6x^2+3x+1) is 2/(sqrt(15))arctan((4sqrt(3))/(sqrt(5))x+sqrt(3/5))+C

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integral of 1/(6x^2+3x+1)

\int \frac{1}{6 x ^{2} + 3 x + 1} d x

\frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5}) + C

Solution steps

\int \frac{1}{6 x ^{2} + 3 x + 1} d x

Complete the square

6 x^{2} + 3 x + 1 : 6 (x + \frac{1}{4})^{2} + \frac{5}{8}

= \int \frac{1}{6 ( x + \frac{1}{4} ) ^{2} + \frac{5}{8}} d x

Apply u-substitution

: \int \frac{8}{48 u ^{2} + 5} d u

= \int \frac{8}{48 u ^{2} + 5} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{48 u ^{2} + 5} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{4 15 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4 15} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot \frac{1}{4 15} arctan (v)

Substitute back

= 8 \cdot \frac{1}{4 15} arctan (\frac{4 3}{5} (x + \frac{1}{4}))

Simplify

8 \cdot \frac{1}{4 15} arctan (\frac{4 3}{5} (x + \frac{1}{4})) : \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5})

= \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5})

Add a constant to the solution

= \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5}) + C

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What is the integral of 1/(6x^2+3x+1) ?
The integral of 1/(6x^2+3x+1) is 2/(sqrt(15))arctan((4sqrt(3))/(sqrt(5))x+sqrt(3/5))+C