integral of (x^2+2x+4)/((x+1)^3)

解

\int \frac{x ^{2} + 2 x + 4}{( x + 1 ) ^{3}} d x

\frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{1}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{1}{( x + 1 ) ^{2}} + C

解答ステップ

\int \frac{x ^{2} + 2 x + 4}{( x + 1 ) ^{3}} d x

拡張

\frac{x ^{2} + 2 x + 4}{( x + 1 ) ^{3}} : \frac{x ^{2}}{( x + 1 ) ^{3}} + \frac{2 x}{( x + 1 ) ^{3}} + \frac{4}{( x + 1 ) ^{3}}

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

\int \frac{x ^{2}}{( x + 1 ) ^{3}} d x = \frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}}

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

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

= \frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - \frac{2}{( x + 1 ) ^{2}}

簡素化

\frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{3}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}} - \frac{2}{( x + 1 ) ^{2}} : \frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{1}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{1}{( x + 1 ) ^{2}}

= \frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{1}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{1}{( x + 1 ) ^{2}}

定数を解答に追加する

= \frac{x ^{3}}{( x + 1 ) ^{3}} + ln ∣ x + 1 ∣ + \frac{1}{x + 1} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{1}{3 ( x + 1 ) ^{3}} + \frac{2 x ^{2}}{3 ( x + 1 ) ^{3}} - \frac{2 ( - 2 x - 1 )}{3 ( x + 1 ) ^{2}} - \frac{1}{( x + 1 ) ^{2}} + C