What is the integral of (x^2-6x+13)^{-1} ?

The integral of (x^2-6x+13)^{-1} is 1/2 arctan((x-3)/2)+C

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integral of (x^2-6x+13)^{-1}

\int (x^{2} - 6 x + 13)^{- 1} d x

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

Solution steps

\int (x^{2} - 6 x + 13)^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{x ^{2} - 6 x + 13} d x

Complete the square

x^{2} - 6 x + 13 : (x - 3)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Substitute back

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

Add a constant to the solution

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

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What is the integral of (x^2-6x+13)^{-1} ?
The integral of (x^2-6x+13)^{-1} is 1/2 arctan((x-3)/2)+C