integral of (2t-1)/(t^2-t+2)

解

\int \frac{2 t - 1}{t ^{2} - t + 2} d t

ln 4 t^{2} - 4 t + 8 + C

解答ステップ

\int \frac{2 t - 1}{t ^{2} - t + 2} d t

拡張

\frac{2 t - 1}{t ^{2} - t + 2} : \frac{2 t}{t ^{2} - t + 2} - \frac{1}{t ^{2} - t + 2}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{2 t}{t ^{2} - t + 2} d t = ln 4 t^{2} - 4 t + 8 + \frac{2}{7} arctan (\frac{2 t - 1}{7})

\int \frac{1}{t ^{2} - t + 2} d t = \frac{2}{7} arctan (\frac{2 t - 1}{7})

= ln 4 t^{2} - 4 t + 8 + \frac{2}{7} arctan (\frac{2 t - 1}{7}) - \frac{2}{7} arctan (\frac{2 t - 1}{7})

類似した元を足す:

\frac{2}{7} arctan (\frac{2 t - 1}{7}) - \frac{2}{7} arctan (\frac{2 t - 1}{7}) = 0

= ln 4 t^{2} - 4 t + 8

定数を解答に追加する

= ln 4 t^{2} - 4 t + 8 + C