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

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

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

\int \frac{( r - 1 )}{r ^{2} - 4 r + 3} d r

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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