What is the integral of 1/(x^2-7x+12) ?

The integral of 1/(x^2-7x+12) is -ln|x-3|+ln|x-4|+C

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integral of 1/(x^2-7x+12)

\int \frac{1}{x ^{2} - 7 x + 12} d x

- ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 7 x + 12} d x

Take the partial fraction of

\frac{1}{x ^{2} - 7 x + 12} : - \frac{1}{x - 3} + \frac{1}{x - 4}

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

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

\int \frac{1}{x - 4} d x = ln ∣ x - 4 ∣

= - ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣

Add a constant to the solution

= - ln ∣ x - 3 ∣ + ln ∣ x - 4 ∣ + C

x y^{2} \frac{d y}{d x} = y^{3} - 3 x^{3}, y (1) = 5

y^{^{'}} = 0.1 \cdot y - 250000

f (x) = 3 cos^{2} (x)

\int 5 cos (π x - 3) d x

\frac{4 x}{x ^{2} + 2}

What is the integral of 1/(x^2-7x+12) ?
The integral of 1/(x^2-7x+12) is -ln|x-3|+ln|x-4|+C