integral of ((x-1)^3)/((x+2)^5)

解

\int \frac{( x - 1 ) ^{3}}{( x + 2 ) ^{5}} d x

- \frac{1}{x + 2} + \frac{9}{2 ( x + 2 ) ^{2}} - \frac{9}{( x + 2 ) ^{3}} - 27 (\frac{x}{( x + 2 ) ^{5}} - \frac{5}{4 ( x + 2 ) ^{4}} + \frac{2}{( x + 2 ) ^{5}}) + C

解答ステップ

\int \frac{( x - 1 ) ^{3}}{( x + 2 ) ^{5}} d x

以下の部分分数を得る:

\frac{( x - 1 ) ^{3}}{( x + 2 ) ^{5}} : \frac{1}{( x + 2 ) ^{2}} - \frac{9}{( x + 2 ) ^{3}} + \frac{27}{( x + 2 ) ^{4}} - \frac{27}{( x + 2 ) ^{5}}

= \int \frac{1}{( x + 2 ) ^{2}} - \frac{9}{( x + 2 ) ^{3}} + \frac{27}{( x + 2 ) ^{4}} - \frac{27}{( x + 2 ) ^{5}} d x

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{1}{( x + 2 ) ^{2}} d x - \int \frac{9}{( x + 2 ) ^{3}} d x + \int \frac{27}{( x + 2 ) ^{4}} d x - \int \frac{27}{( x + 2 ) ^{5}} d x

\int \frac{1}{( x + 2 ) ^{2}} d x = - \frac{1}{x + 2}

\int \frac{9}{( x + 2 ) ^{3}} d x = - \frac{9}{2 ( x + 2 ) ^{2}}

\int \frac{27}{( x + 2 ) ^{4}} d x = - \frac{9}{( x + 2 ) ^{3}}

\int \frac{27}{( x + 2 ) ^{5}} d x = 27 (\frac{x}{( x + 2 ) ^{5}} - \frac{5}{4 ( x + 2 ) ^{4}} + \frac{2}{( x + 2 ) ^{5}})

= - \frac{1}{x + 2} - (- \frac{9}{2 ( x + 2 ) ^{2}}) - \frac{9}{( x + 2 ) ^{3}} - 27 (\frac{x}{( x + 2 ) ^{5}} - \frac{5}{4 ( x + 2 ) ^{4}} + \frac{2}{( x + 2 ) ^{5}})

簡素化

= - \frac{1}{x + 2} + \frac{9}{2 ( x + 2 ) ^{2}} - \frac{9}{( x + 2 ) ^{3}} - 27 (\frac{x}{( x + 2 ) ^{5}} - \frac{5}{4 ( x + 2 ) ^{4}} + \frac{2}{( x + 2 ) ^{5}})

定数を解答に追加する

= - \frac{1}{x + 2} + \frac{9}{2 ( x + 2 ) ^{2}} - \frac{9}{( x + 2 ) ^{3}} - 27 (\frac{x}{( x + 2 ) ^{5}} - \frac{5}{4 ( x + 2 ) ^{4}} + \frac{2}{( x + 2 ) ^{5}}) + C