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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} - 4 x + 4} d x

Apply Integration By Parts

: - \frac{x}{x - 2} - \int - \frac{1}{x - 2} d x

= 3 (- \frac{x}{x - 2} - \int - \frac{1}{x - 2} d x)

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

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

Simplify

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

Add a constant to the solution

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

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