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

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

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

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

\frac{4}{3} arctan (23 x + 3) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

= 2 \cdot \int \frac{1}{3 ( x + \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

Apply u-substitution

: \int \frac{4}{12 u ^{2} + 1} d u

= 2 \cdot \int \frac{4}{12 u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot 4 \cdot \int \frac{1}{12 u ^{2} + 1} d u

Apply Integral Substitution

= 2 \cdot 4 \cdot \int \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot 4 \cdot \frac{1}{2 3} arctan (v)

Substitute back

= 2 \cdot 4 \cdot \frac{1}{2 3} arctan (23 (x + \frac{1}{2}))

Simplify

2 \cdot 4 \cdot \frac{1}{2 3} arctan (23 (x + \frac{1}{2})) : \frac{4}{3} arctan (23 x + 3)

= \frac{4}{3} arctan (23 x + 3)

Add a constant to the solution

= \frac{4}{3} arctan (23 x + 3) + C

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