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

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

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

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

\frac{1}{5} arctan (\frac{2 x - 1}{5}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2}{4 u ^{2} + 5} d u

= \int \frac{2}{4 u ^{2} + 5} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 u ^{2} + 5} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 5 ( 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 \frac{1}{2 5} \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 \frac{1}{2 5} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2 5} arctan (\frac{2}{5} (x - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2 5} arctan (\frac{2}{5} (x - \frac{1}{2})) : \frac{1}{5} arctan (\frac{2 x - 1}{5})

= \frac{1}{5} arctan (\frac{2 x - 1}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{2 x - 1}{5}) + C

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