What is the integral of (2z-3)/(z^2-6z+13) ?

The integral of (2z-3)/(z^2-6z+13) is ln|z^2-6z+13|+3/2 arctan(1/2 (z-3))+C

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integral of (2z-3)/(z^2-6z+13)

\int \frac{2 z - 3}{z ^{2} - 6 z + 13} d z

ln z^{2} - 6 z + 13 + \frac{3}{2} arctan (\frac{1}{2} (z - 3)) + C

Solution steps

\int \frac{2 z - 3}{z ^{2} - 6 z + 13} d z

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u + 3}{u ^{2} + 4} d u

= \int \frac{2 u + 3}{u ^{2} + 4} d u

Expand

\frac{2 u + 3}{u ^{2} + 4} : \frac{2 u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4}

Apply the Sum Rule:

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

= \int \frac{2 u}{u ^{2} + 4} d u + \int \frac{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= ln u^{2} + 4 + \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = z - 3

= ln (z - 3)^{2} + 4 + \frac{3}{2} arctan (\frac{z - 3}{2})

Simplify

ln (z - 3)^{2} + 4 + \frac{3}{2} arctan (\frac{z - 3}{2}) : ln z^{2} - 6 z + 13 + \frac{3}{2} arctan (\frac{1}{2} (z - 3))

= ln z^{2} - 6 z + 13 + \frac{3}{2} arctan (\frac{1}{2} (z - 3))

Add a constant to the solution

= ln z^{2} - 6 z + 13 + \frac{3}{2} arctan (\frac{1}{2} (z - 3)) + C

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What is the integral of (2z-3)/(z^2-6z+13) ?
The integral of (2z-3)/(z^2-6z+13) is ln|z^2-6z+13|+3/2 arctan(1/2 (z-3))+C