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

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

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

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

ln x^{2} - 4 x + 5 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{2} - 4 x + 5

= ln x^{2} - 4 x + 5

Add a constant to the solution

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

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