integral of (16s+16)/((s^2+1)(s-1)^3)

解

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}} + C

解答ステップ

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

以下の部分分数を得る:

\frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{8}{s ^{2} + 1} - \frac{8}{( s - 1 ) ^{2}} + \frac{16}{( s - 1 ) ^{3}}

= \int \frac{8}{s ^{2} + 1} - \frac{8}{( s - 1 ) ^{2}} + \frac{16}{( 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{8}{s ^{2} + 1} d s - \int \frac{8}{( s - 1 ) ^{2}} d s + \int \frac{16}{( s - 1 ) ^{3}} d s

\int \frac{8}{s ^{2} + 1} d s = 8 arctan (s)

\int \frac{8}{( s - 1 ) ^{2}} d s = - \frac{8}{s - 1}

\int \frac{16}{( s - 1 ) ^{3}} d s = - \frac{8}{( s - 1 ) ^{2}}

= 8 arctan (s) - (- \frac{8}{s - 1}) - \frac{8}{( s - 1 ) ^{2}}

簡素化

= 8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}}

定数を解答に追加する

= 8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}} + C