integral of 3/((t^2-9)^2)

解

\int \frac{3}{( t ^{2} - 9 ) ^{2}} d t

3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )}) + C

解答ステップ

\int \frac{3}{( t ^{2} - 9 ) ^{2}} d t

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{( t ^{2} - 9 ) ^{2}} d t

以下の部分分数を得る:

\frac{1}{( t ^{2} - 9 ) ^{2}} : \frac{1}{108 ( t + 3 )} + \frac{1}{36 ( t + 3 ) ^{2}} - \frac{1}{108 ( t - 3 )} + \frac{1}{36 ( t - 3 ) ^{2}}

= 3 \cdot \int \frac{1}{108 ( t + 3 )} + \frac{1}{36 ( t + 3 ) ^{2}} - \frac{1}{108 ( t - 3 )} + \frac{1}{36 ( 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

= 3 (\int \frac{1}{108 ( t + 3 )} d t + \int \frac{1}{36 ( t + 3 ) ^{2}} d t - \int \frac{1}{108 ( t - 3 )} d t + \int \frac{1}{36 ( t - 3 ) ^{2}} d t)

\int \frac{1}{108 ( t + 3 )} d t = \frac{1}{108} ln ∣ t + 3 ∣

\int \frac{1}{36 ( t + 3 ) ^{2}} d t = - \frac{1}{36 ( t + 3 )}

\int \frac{1}{108 ( t - 3 )} d t = \frac{1}{108} ln ∣ t - 3 ∣

\int \frac{1}{36 ( t - 3 ) ^{2}} d t = - \frac{1}{36 ( t - 3 )}

= 3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )})

定数を解答に追加する

= 3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )}) + C