integral of (2x+2)/(x^2-4x+9)

解

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

\frac{6}{5} arctan (\frac{1}{5} (x - 2)) + ln \frac{1}{5} (x - 2)^{2} + 1 + C

解答ステップ

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

部分積分を適用する

= \frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - \int \frac{2}{5} arctan (\frac{1}{5} (x - 2)) d x

\int \frac{2}{5} arctan (\frac{1}{5} (x - 2)) d x = \frac{2}{5} x arctan (\frac{1}{5} (x - 2)) - ln \frac{1}{5} (x - 2)^{2} + 1 - \frac{4}{5} arctan (\frac{1}{5} (x - 2))

= \frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - (\frac{2}{5} x arctan (\frac{1}{5} (x - 2)) - ln \frac{1}{5} (x - 2)^{2} + 1 - \frac{4}{5} arctan (\frac{1}{5} (x - 2)))

簡素化

\frac{1}{5} arctan (\frac{1}{5} (x - 2)) (2 x + 2) - (\frac{2}{5} x arctan (\frac{1}{5} (x - 2)) - ln \frac{1}{5} (x - 2)^{2} + 1 - \frac{4}{5} arctan (\frac{1}{5} (x - 2))) : \frac{6}{5} arctan (\frac{1}{5} (x - 2)) + ln \frac{1}{5} (x - 2)^{2} + 1

= \frac{6}{5} arctan (\frac{1}{5} (x - 2)) + ln \frac{1}{5} (x - 2)^{2} + 1

定数を解答に追加する

= \frac{6}{5} arctan (\frac{1}{5} (x - 2)) + ln \frac{1}{5} (x - 2)^{2} + 1 + C