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

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

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

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

ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

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

Complete the square

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

= \int \frac{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

\int_{0}^{3} 2^{x} d x

n = 0 \sum \infty \frac{n ^{3}}{2 ^{n}}

\int \frac{1}{sin ^{2} ( a ) x} d x

5 x - 2 x

y^{^{'}} = x^{5} + 2 x^{4} - x - 2

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