integral of (14s+14)/((s^2+1)(s-1)^3)

解

\int \frac{14 s + 14}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

7 arctan (s) + \frac{7}{s - 1} - \frac{7}{( s - 1 ) ^{2}} + C

解答ステップ

\int \frac{14 s + 14}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

以下の部分分数を得る:

\frac{14 s + 14}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{7}{s ^{2} + 1} - \frac{7}{( s - 1 ) ^{2}} + \frac{14}{( s - 1 ) ^{3}}

= \int \frac{7}{s ^{2} + 1} - \frac{7}{( s - 1 ) ^{2}} + \frac{14}{( s - 1 ) ^{3}} d s

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{7}{s ^{2} + 1} d s - \int \frac{7}{( s - 1 ) ^{2}} d s + \int \frac{14}{( s - 1 ) ^{3}} d s

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

\int \frac{7}{( s - 1 ) ^{2}} d s = - \frac{7}{s - 1}

\int \frac{14}{( s - 1 ) ^{3}} d s = - \frac{7}{( s - 1 ) ^{2}}

= 7 arctan (s) - (- \frac{7}{s - 1}) - \frac{7}{( s - 1 ) ^{2}}

簡素化

= 7 arctan (s) + \frac{7}{s - 1} - \frac{7}{( s - 1 ) ^{2}}

定数を解答に追加する

= 7 arctan (s) + \frac{7}{s - 1} - \frac{7}{( s - 1 ) ^{2}} + C