What is the integral of (x-3)/(x^2-6x+10) ?

The integral of (x-3)/(x^2-6x+10) is 1/2 ln|x^2-6x+10|+C

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integral of (x-3)/(x^2-6x+10)

\int \frac{x - 3}{x ^{2} - 6 x + 10} d x

\frac{1}{2} ln x^{2} - 6 x + 10 + C

Solution steps

\int \frac{x - 3}{x ^{2} - 6 x + 10} d x

Apply u-substitution

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

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

Apply u-substitution

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

= \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 - 3)^{2} + 1

Simplify

\frac{1}{2} ln (x - 3)^{2} + 1 : \frac{1}{2} ln x^{2} - 6 x + 10

= \frac{1}{2} ln x^{2} - 6 x + 10

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 6 x + 10 + C

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What is the integral of (x-3)/(x^2-6x+10) ?
The integral of (x-3)/(x^2-6x+10) is 1/2 ln|x^2-6x+10|+C