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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{4}^{x} \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 )}

= 4 \cdot \int_{4}^{x} - \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

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

\int_{4}^{x} \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} (ln ∣ x - 1 ∣ - ln (3))

\int_{4}^{x} \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

= 4 (- \frac{1}{2} (ln ∣ x - 1 ∣ - ln (3)) + \frac{1}{2} ln ∣ x - 3 ∣)

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