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

解

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

\frac{1}{8} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{17}{8} ln ∣ x + 3 ∣ + \frac{5}{4 ( x + 3 )} + \frac{1}{4 ( x + 3 ) ^{2}} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x + 1 ) ( x + 2 ) ^{2} ( x + 3 ) ^{3}} : \frac{1}{8 ( x + 1 )} + \frac{2}{x + 2} - \frac{1}{( x + 2 ) ^{2}} - \frac{17}{8 ( x + 3 )} - \frac{5}{4 ( x + 3 ) ^{2}} - \frac{1}{2 ( x + 3 ) ^{3}}

= \int \frac{1}{8 ( x + 1 )} + \frac{2}{x + 2} - \frac{1}{( x + 2 ) ^{2}} - \frac{17}{8 ( x + 3 )} - \frac{5}{4 ( x + 3 ) ^{2}} - \frac{1}{2 ( x + 3 ) ^{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{1}{8 ( x + 1 )} d x + \int \frac{2}{x + 2} d x - \int \frac{1}{( x + 2 ) ^{2}} d x - \int \frac{17}{8 ( x + 3 )} d x - \int \frac{5}{4 ( x + 3 ) ^{2}} d x - \int \frac{1}{2 ( x + 3 ) ^{3}} d x

\int \frac{1}{8 ( x + 1 )} d x = \frac{1}{8} ln ∣ x + 1 ∣

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

\int \frac{17}{8 ( x + 3 )} d x = \frac{17}{8} ln ∣ x + 3 ∣

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

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

= \frac{1}{8} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - (- \frac{1}{x + 2}) - \frac{17}{8} ln ∣ x + 3 ∣ - (- \frac{5}{4 ( x + 3 )}) - (- \frac{1}{4 ( x + 3 ) ^{2}})

簡素化

= \frac{1}{8} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{17}{8} ln ∣ x + 3 ∣ + \frac{5}{4 ( x + 3 )} + \frac{1}{4 ( x + 3 ) ^{2}}

定数を解答に追加する

= \frac{1}{8} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ + \frac{1}{x + 2} - \frac{17}{8} ln ∣ x + 3 ∣ + \frac{5}{4 ( x + 3 )} + \frac{1}{4 ( x + 3 ) ^{2}} + C