What is the integral of (-x+4)/(x^2-2x+4) ?

The integral of (-x+4)/(x^2-2x+4) is -1/2 ln| 1/3 (x-1)^2+1|+sqrt(3)arctan(1/(sqrt(3))(x-1))+C

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integral of (-x+4)/(x^2-2x+4)

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

- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1)) + C

Solution steps

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

Complete the square

x^{2} - 2 x + 4 : (x - 1)^{2} + 3

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

Apply u-substitution

: \int \frac{- u + 3}{u ^{2} + 3} d u

= \int \frac{- u + 3}{u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{- v + 3}{v ^{2} + 1} d v

Expand

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{v ^{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{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v

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

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= - \frac{1}{2} ln v^{2} + 1 + 3 arctan (v)

Substitute back

= - \frac{1}{2} ln (\frac{x - 1}{3})^{2} + 1 + 3 arctan (\frac{x - 1}{3})

Simplify

- \frac{1}{2} ln (\frac{x - 1}{3})^{2} + 1 + 3 arctan (\frac{x - 1}{3}) : - \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))

= - \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))

Add a constant to the solution

= - \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1)) + C

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What is the integral of (-x+4)/(x^2-2x+4) ?
The integral of (-x+4)/(x^2-2x+4) is -1/2 ln| 1/3 (x-1)^2+1|+sqrt(3)arctan(1/(sqrt(3))(x-1))+C