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

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

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

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

\frac{1}{5} arctan (\frac{3 x - 1}{5}) + C

Solution steps

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

Complete the square

3 x^{2} - 2 x + 2 : 3 (x - \frac{1}{3})^{2} + \frac{5}{3}

= \int \frac{1}{3 ( x - \frac{1}{3} ) ^{2} + \frac{5}{3}} d x

Apply u-substitution

: \int \frac{3}{9 u ^{2} + 5} d u

= \int \frac{3}{9 u ^{2} + 5} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} + 5} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{3 5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{3 5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{3 5} arctan (v)

Substitute back

= 3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (x - \frac{1}{3}))

Simplify

3 \cdot \frac{1}{3 5} arctan (\frac{3}{5} (x - \frac{1}{3})) : \frac{1}{5} arctan (\frac{3 x - 1}{5})

= \frac{1}{5} arctan (\frac{3 x - 1}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{3 x - 1}{5}) + C

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