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

解

\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

解答ステップ

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

部分積分を適用する

= 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))

簡素化

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

定数を解答に追加する

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