What is the integral of (2x-3)/(x^2-4x+5) ?

The integral of (2x-3)/(x^2-4x+5) is ln|x^2-4x+5|+arctan(x-2)+C

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integral of (2x-3)/(x^2-4x+5)

\int \frac{2 x - 3}{x ^{2} - 4 x + 5} d x

ln x^{2} - 4 x + 5 + arctan (x - 2) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} - 4 x + 5} d x

Complete the square

x^{2} - 4 x + 5 : (x - 2)^{2} + 1

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

Apply u-substitution

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

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

Expand

\frac{2 u + 1}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

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

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

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

= ln u^{2} + 1 + arctan (u)

Substitute back

u = x - 2

= ln (x - 2)^{2} + 1 + arctan (x - 2)

Expand

(x - 2)^{2} + 1 : x^{2} - 4 x + 5

= ln x^{2} - 4 x + 5 + arctan (x - 2)

Add a constant to the solution

= ln x^{2} - 4 x + 5 + arctan (x - 2) + C

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What is the integral of (2x-3)/(x^2-4x+5) ?
The integral of (2x-3)/(x^2-4x+5) is ln|x^2-4x+5|+arctan(x-2)+C