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

y = 3 x + 6 cos (x), (0, 6)

\int \frac{1}{2 - x ^{\frac{1}{5}}} d x

f (x) = \frac{x + x e ^{x}}{x ^{2} - 5}

\frac{d}{d x} (3 1 + tan (x^{2}))

x (3 - x)^{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