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)

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

- \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

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

\frac{d}{d x} (8 x^{2} + 7)

f (x) = (2 x^{3} + 2) (x^{4} - 3 x)

\int_{0}^{2} 3^{- x} d x

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{x} cos (3 x)

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