What is the integral of (x-4)/(x^2-4x+8) ?

The integral of (x-4)/(x^2-4x+8) is 1/2 ln|((x-2)^2)/4+1|-arctan(1/2 (x-2))+C

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integral of (x-4)/(x^2-4x+8)

\int \frac{x - 4}{x ^{2} - 4 x + 8} d x

\frac{1}{2} ln \frac{( x - 2 ) ^{2}}{4} + 1 - arctan (\frac{1}{2} (x - 2)) + C

Solution steps

\int \frac{x - 4}{x ^{2} - 4 x + 8} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 4 u + 8} d u

= \int \frac{u}{u ^{2} + 4 u + 8} d u

Complete the square

u^{2} + 4 u + 8 : (u + 2)^{2} + 4

= \int \frac{u}{( u + 2 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v - 2}{v ^{2} + 4} d v

= \int \frac{v - 2}{v ^{2} + 4} d v

Apply Integral Substitution

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

Expand

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

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

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

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

Substitute back

= \frac{1}{2} ln (\frac{x - 4 + 2}{2})^{2} + 1 - arctan (\frac{x - 4 + 2}{2})

Simplify

\frac{1}{2} ln (\frac{x - 4 + 2}{2})^{2} + 1 - arctan (\frac{x - 4 + 2}{2}) : \frac{1}{2} ln \frac{( x - 2 ) ^{2}}{4} + 1 - arctan (\frac{1}{2} (x - 2))

= \frac{1}{2} ln \frac{( x - 2 ) ^{2}}{4} + 1 - arctan (\frac{1}{2} (x - 2))

Add a constant to the solution

= \frac{1}{2} ln \frac{( x - 2 ) ^{2}}{4} + 1 - arctan (\frac{1}{2} (x - 2)) + C

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What is the integral of (x-4)/(x^2-4x+8) ?
The integral of (x-4)/(x^2-4x+8) is 1/2 ln|((x-2)^2)/4+1|-arctan(1/2 (x-2))+C