What is the integral of (2-x)/(x^2-2x+1) ?

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

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integral of (2-x)/(x^2-2x+1)

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

- ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

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

Complete the square

x^{2} - 2 x + 1 : (x - 1)^{2}

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

Apply u-substitution

: \int \frac{- u + 1}{u ^{2}} d u

= \int \frac{- u + 1}{u ^{2}} d u

Expand

\frac{- u + 1}{u ^{2}} : - \frac{1}{u} + \frac{1}{u ^{2}}

= \int - \frac{1}{u} + \frac{1}{u ^{2}} d u

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}{u} d u + \int \frac{1}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= - ln ∣ u ∣ - \frac{1}{u}

Substitute back

u = x - 1

= - ln ∣ x - 1 ∣ - \frac{1}{x - 1}

Add a constant to the solution

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

x^{3}, x

7 x^{3} e^{x}

f (x) = - \frac{1}{( x - 5 ) ^{2}}, at x = - 4

\int \frac{x ^{3} + 3 x - 1}{x ^{4} - 4 x ^{2}} d x

x \to 0 - lim (\frac{e ^{3 x}}{2 x})

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