What is the integral of (x-1)/(x^2-2x+5) ?

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

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integral of (x-1)/(x^2-2x+5)

\int \frac{x - 1}{x ^{2} - 2 x + 5} d x

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

Solution steps

\int \frac{x - 1}{x ^{2} - 2 x + 5} d x

Apply u-substitution

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

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

\int \frac{x - 1}{9 x ^{2} - 18 x + 17} d x

\int \frac{5 x}{9 - x ^{2}} d x

\int \frac{1}{2 x ^{2} - 8} d x

5^{x}

\int x^{2} x^{3} - 2 d x

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