What is the integral of 1/(4x^2-4x+10) ?

The integral of 1/(4x^2-4x+10) is 1/6 arctan(2/3 x-1/3)+C

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integral of 1/(4x^2-4x+10)

\int \frac{1}{4 x ^{2} - 4 x + 10} d x

\frac{1}{6} arctan (\frac{2}{3} x - \frac{1}{3}) + C

Solution steps

\int \frac{1}{4 x ^{2} - 4 x + 10} d x

Complete the square

4 x^{2} - 4 x + 10 : 4 (x - \frac{1}{2})^{2} + 9

= \int \frac{1}{4 ( x - \frac{1}{2} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 9} d u

= \int \frac{1}{4 u ^{2} + 9} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{2}{3} (x - \frac{1}{2}))

Expand

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

= \frac{1}{6} arctan (\frac{2}{3} x - \frac{1}{3})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{2}{3} x - \frac{1}{3}) + C

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What is the integral of 1/(4x^2-4x+10) ?
The integral of 1/(4x^2-4x+10) is 1/6 arctan(2/3 x-1/3)+C