integral of (t^2-2)/((t^2-t-6)^2)

解

\int \frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} d t

- \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )} + C

解答ステップ

\int \frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} d t

以下の部分分数を得る:

\frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} : - \frac{16}{125 ( t + 2 )} + \frac{2}{25 ( t + 2 ) ^{2}} + \frac{16}{125 ( t - 3 )} + \frac{7}{25 ( t - 3 ) ^{2}}

= \int - \frac{16}{125 ( t + 2 )} + \frac{2}{25 ( t + 2 ) ^{2}} + \frac{16}{125 ( t - 3 )} + \frac{7}{25 ( t - 3 ) ^{2}} d t

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{16}{125 ( t + 2 )} d t + \int \frac{2}{25 ( t + 2 ) ^{2}} d t + \int \frac{16}{125 ( t - 3 )} d t + \int \frac{7}{25 ( t - 3 ) ^{2}} d t

\int \frac{16}{125 ( t + 2 )} d t = \frac{16}{125} ln ∣ t + 2 ∣

\int \frac{2}{25 ( t + 2 ) ^{2}} d t = - \frac{2}{25 ( t + 2 )}

\int \frac{16}{125 ( t - 3 )} d t = \frac{16}{125} ln ∣ t - 3 ∣

\int \frac{7}{25 ( t - 3 ) ^{2}} d t = - \frac{7}{25 ( t - 3 )}

= - \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )}

定数を解答に追加する

= - \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )} + C