What is the integral of 1/(sqrt(x^2-4x+3)) ?

The integral of 1/(sqrt(x^2-4x+3)) is ln|sqrt(x^2-4x+3)+x-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x^2-4x+3))

\int \frac{1}{x ^{2} - 4 x + 3} d x

ln x^{2} - 4 x + 3 + x - 2 + C

Solution steps

\int \frac{1}{x ^{2} - 4 x + 3} d x

Complete the square

x^{2} - 4 x + 3 : (x - 2)^{2} - 1

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ : ln x^{2} - 4 x + 3 + x - 2

= ln x^{2} - 4 x + 3 + x - 2

Add a constant to the solution

= ln x^{2} - 4 x + 3 + x - 2 + C

\frac{d}{d y} (\frac{x ^{3} y - x y ^{3}}{x ^{2} + y ^{2}})

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

\int \frac{1}{2 x ^{2} - 4 x + 14} d x

\int x 5^{x} d x

\frac{d}{d x} (e e^{x} e^{x^{2}})

What is the integral of 1/(sqrt(x^2-4x+3)) ?
The integral of 1/(sqrt(x^2-4x+3)) is ln|sqrt(x^2-4x+3)+x-2|+C