What is the integral of (2x-3)/(x^2-3x+8) ?

The integral of (2x-3)/(x^2-3x+8) is ln|x^2-3x+8|+C

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integral of (2x-3)/(x^2-3x+8)

\int \frac{2 x - 3}{x ^{2} - 3 x + 8} d x

ln x^{2} - 3 x + 8 + C

Solution steps

\int \frac{2 x - 3}{x ^{2} - 3 x + 8} d x

Apply u-substitution

= \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 + 8

= ln x^{2} - 3 x + 8

Add a constant to the solution

= ln x^{2} - 3 x + 8 + C

\int \frac{1 + ln ( x )}{( x ln ( x ) ) 2} d x

\int \frac{e ^{3 x}}{1 - e ^{3 x}} d x

\int \frac{x ^{2} + 1}{x ^{2} - 4} d x

\int 400 e^{- 0.1 x} + 8 d x

\int x^{4} e (1 - x^{5}) d x

What is the integral of (2x-3)/(x^2-3x+8) ?
The integral of (2x-3)/(x^2-3x+8) is ln|x^2-3x+8|+C