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

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

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

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

ln x^{2} - 2 x + 2 + 2 arctan (x - 1) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 (\frac{1}{2} ln u^{2} + 1 + arctan (u))

Substitute back

u = x - 1

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

Simplify

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

= ln x^{2} - 2 x + 2 + 2 arctan (x - 1)

Add a constant to the solution

= ln x^{2} - 2 x + 2 + 2 arctan (x - 1) + C

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