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

The integral of ((2x+2))/((x^2-4x+8)) is 3arctan(1/2 x-1)+ln| 1/4 x^2-x+2|+C

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

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

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

Solution steps

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

Apply Integration By Parts

: arctan (\frac{1}{2} (x - 2)) (x + 1) - \int arctan (\frac{1}{2} (x - 2)) d x

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

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

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

Simplify

arctan (\frac{1}{2} (x - 2)) (x + 1) - (x arctan (\frac{1}{2} x - 1) - ln \frac{1}{4} x^{2} - x + 2 - 2 arctan (\frac{1}{2} x - 1)) : 3 arctan (\frac{1}{2} x - 1) + ln \frac{1}{4} x^{2} - x + 2

= 3 arctan (\frac{1}{2} x - 1) + ln \frac{1}{4} x^{2} - x + 2

Add a constant to the solution

= 3 arctan (\frac{1}{2} x - 1) + ln \frac{1}{4} x^{2} - x + 2 + C

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What is the integral of ((2x+2))/((x^2-4x+8)) ?
The integral of ((2x+2))/((x^2-4x+8)) is 3arctan(1/2 x-1)+ln| 1/4 x^2-x+2|+C