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

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

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integral of (2-x)/(x^2-4x)

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

- \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{2 x} - \frac{1}{2 ( 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}{2 x} d x - \int \frac{1}{2 ( x - 4 )} d x

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

\int \frac{1}{2 ( x - 4 )} d x = \frac{1}{2} ln ∣ x - 4 ∣

= - \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 4 ∣ + C

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\int (x - 1) csc^{2} (2 x - x^{2}) d x

\int (x + 4) e^{x} d x

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