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

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

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

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

ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{x - 1} : ln ∣ x - 1 ∣ - \frac{2}{x - 1}

= ln ∣ x - 1 ∣ - \frac{2}{x - 1}

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

\frac{( x ^{2} - 1 )}{x ^{2} + x + 1}

2 x y y^{^{'}} = x^{2} + 2 y^{2}

f (x) = 2 e^{4 x}

x^{2} + 8, - 2, 1

x y^{^{'}} + 2 y = 2 sin (x)

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