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

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

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

integral

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \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

= \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)

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{cos ^{23} ( x )}{sin ( x )}

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

2 x (1 - x^{- 3})

\frac{3}{4} - \frac{x}{4}

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