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

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

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

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

- \frac{x}{x - 2} + ln ∣ x - 2 ∣ + C

Solution steps

\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

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

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

Simplify

= - \frac{x}{x - 2} + ln ∣ x - 2 ∣

Add a constant to the solution

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

\int + 4 8 - 3 x d x

t an g e n t 4 x^{3} - 3 x^{2} + 4, at x = 2

\frac{d y}{d x} = 2 y - x^{2}

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

\frac{\partial}{\partial x} (\frac{5}{e ^{x}})

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