What is the integral of (7-x)/(x^2-4x+4) ?

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

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integral of (7-x)/(x^2-4x+4)

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

- \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

Expand

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

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

\int \frac{7}{x ^{2} - 4 x + 4} d x = - \frac{7}{x - 2}

\int \frac{x}{x ^{2} - 4 x + 4} d x = - \frac{x}{x - 2} + ln ∣ x - 2 ∣

= - \frac{7}{x - 2} - (- \frac{x}{x - 2} + ln ∣ x - 2 ∣)

Simplify

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{7}{x - 2} + \frac{x}{x - 2} - ln ∣ x - 2 ∣ + C

\frac{d}{d x} (cos (w x))

x \to - \infty lim (- 3 x - 2)

f (x) = x - e

\frac{d}{d t} (4 \frac{e ^{5 t}}{t ^{2} + 1})

\frac{y ^{^{''}}}{y} = - 3

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