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

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

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

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

ln x^{2} - 3 x + 6 + C

Solution steps

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

= ln x^{2} - 3 x + 6

Add a constant to the solution

= ln x^{2} - 3 x + 6 + C

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

\int - 2 e^{- t} + 8 t + 6 d t

\int x_{0} d x

\int \frac{1}{5 x ^{2} - 10 x + 6} d x

\int sec^{2} (t) tan (t) d t

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