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

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

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

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

f (x) = - 2 x^{2}

n = 0 \sum \infty \frac{n}{n + 1} x^{n}

\frac{d}{d x} (5 e^{- 2 x})

\int - 4 cos^{2} (x) d x

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

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