integral of 5/((t^2-25)^2)

解

\int \frac{5}{( t ^{2} - 25 ) ^{2}} d t

5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )}) + C

解答ステップ

\int \frac{5}{( t ^{2} - 25 ) ^{2}} d t

定数を除く:

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

= 5 \cdot \int \frac{1}{( t ^{2} - 25 ) ^{2}} d t

以下の部分分数を得る:

\frac{1}{( t ^{2} - 25 ) ^{2}} : \frac{1}{500 ( t + 5 )} + \frac{1}{100 ( t + 5 ) ^{2}} - \frac{1}{500 ( t - 5 )} + \frac{1}{100 ( t - 5 ) ^{2}}

= 5 \cdot \int \frac{1}{500 ( t + 5 )} + \frac{1}{100 ( t + 5 ) ^{2}} - \frac{1}{500 ( t - 5 )} + \frac{1}{100 ( t - 5 ) ^{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

= 5 (\int \frac{1}{500 ( t + 5 )} d t + \int \frac{1}{100 ( t + 5 ) ^{2}} d t - \int \frac{1}{500 ( t - 5 )} d t + \int \frac{1}{100 ( t - 5 ) ^{2}} d t)

\int \frac{1}{500 ( t + 5 )} d t = \frac{1}{500} ln ∣ t + 5 ∣

\int \frac{1}{100 ( t + 5 ) ^{2}} d t = - \frac{1}{100 ( t + 5 )}

\int \frac{1}{500 ( t - 5 )} d t = \frac{1}{500} ln ∣ t - 5 ∣

\int \frac{1}{100 ( t - 5 ) ^{2}} d t = - \frac{1}{100 ( t - 5 )}

= 5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )})

定数を解答に追加する

= 5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )}) + C