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

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

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

integral

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

- \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( 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{1}{2 ( x - 1 )} d x + \int \frac{1}{2 ( x - 3 )} d x

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

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

= - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

\frac{1}{( 1 - t ^{2} )}

\frac{x}{( x ^{3} + 1 )}

\frac{1}{( x - x ^{2} )}

3 + 2 x

(\frac{x - 1}{x})^{3}

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