What is the integral of (2x-3)/(x^2-3x+1) ?

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

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integral of (2x-3)/(x^2-3x+1)

\int \frac{2 x - 3}{x ^{2} - 3 x + 1} d x

ln x^{2} - 3 x + 1 + C

Solution steps

\int \frac{2 x - 3}{x ^{2} - 3 x + 1} 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 + 1

= ln x^{2} - 3 x + 1

Add a constant to the solution

= ln x^{2} - 3 x + 1 + C

e^{x + 1}

\int 21 e^{x} d x

\int_{1}^{3} \frac{4}{x ^{2}} d x

9 - 5 x

\frac{d y}{d t} = e^{- 2 t + 3 y}

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