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

The integral of (x-2)/(x^2-4x+5) is 1/2 ln|x^2-4x+5|+C

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

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

\frac{1}{2} ln x^{2} - 4 x + 5 + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Apply u-substitution

= \int \frac{1}{2 v} 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} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

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

= \frac{1}{2} ln x^{2} - 4 x + 5

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 4 x + 5 + C

What is the integral of (x-2)/(x^2-4x+5) ?
The integral of (x-2)/(x^2-4x+5) is 1/2 ln|x^2-4x+5|+C