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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{u ^{2} - 6 u + 8} d u

Complete the square

u^{2} - 6 u + 8 : (u - 3)^{2} - 1

= \int \frac{1}{( u - 3 ) ^{2} - 1} d u

Apply u-substitution

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

= \int \frac{1}{v ^{2} - 1} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= ln ∣ tan (w) + sec (w) ∣

Substitute back

= ln ∣ tan (arcsec (2 x - 3)) + sec (arcsec (2 x - 3)) ∣

Simplify

ln ∣ tan (arcsec (2 x - 3)) + sec (arcsec (2 x - 3)) ∣ : ln 2 x^{2} - 3 x + 2 + 2 x - 3

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

Add a constant to the solution

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

(x^{2} - 1) \frac{d y}{d x} + 2 y = (x + 1)^{2}

f (x) = (2 x + 3)^{2}

\int_{0}^{2} 2 x^{3} x^{2} + 4 d x

x \to 1 lim (x^{3} - 3)

\int arcsin (8 x) d x

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