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

The integral of x/(3x^2+3x+1) is 1/6 ln|12x^2+12x+4|-1/(sqrt(3))arctan(2sqrt(3)x+sqrt(3))+C

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

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

\frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{3} arctan (23 x + 3) + C

Solution steps

\int \frac{x}{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}

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

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{12 u ^{2} + 1} d u

= \int \frac{2 ( 2 u - 1 )}{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 \int \frac{2 u - 1}{12 u ^{2} + 1} d u

Expand

\frac{2 u - 1}{12 u ^{2} + 1} : \frac{2 u}{12 u ^{2} + 1} - \frac{1}{12 u ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{2 u}{12 u ^{2} + 1} d u - \int \frac{1}{12 u ^{2} + 1} d u)

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

\int \frac{1}{12 u ^{2} + 1} d u = \frac{1}{2 3} arctan (23 u)

= 2 (\frac{1}{12} ln 12 u^{2} + 1 - \frac{1}{2 3} arctan (23 u))

Substitute back

u = x + \frac{1}{2}

= 2 (\frac{1}{12} ln 12 (x + \frac{1}{2})^{2} + 1 - \frac{1}{2 3} arctan (23 (x + \frac{1}{2})))

Simplify

2 (\frac{1}{12} ln 12 (x + \frac{1}{2})^{2} + 1 - \frac{1}{2 3} arctan (23 (x + \frac{1}{2}))) : \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{3} arctan (23 x + 3)

= \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{3} arctan (23 x + 3)

Add a constant to the solution

= \frac{1}{6} ln 12 x^{2} + 12 x + 4 - \frac{1}{3} arctan (23 x + 3) + C

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What is the integral of x/(3x^2+3x+1) ?
The integral of x/(3x^2+3x+1) is 1/6 ln|12x^2+12x+4|-1/(sqrt(3))arctan(2sqrt(3)x+sqrt(3))+C