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

解

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

- \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{3} + 4}{( x ^{2} - 1 ) ( x ^{2} + 3 x + 2 )} : - \frac{3}{4 ( x + 1 )} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{5}{12 ( x - 1 )} + \frac{4}{3 ( x + 2 )}

= \int - \frac{3}{4 ( x + 1 )} - \frac{3}{2 ( x + 1 ) ^{2}} + \frac{5}{12 ( x - 1 )} + \frac{4}{3 ( x + 2 )} 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{3}{4 ( x + 1 )} d x - \int \frac{3}{2 ( x + 1 ) ^{2}} d x + \int \frac{5}{12 ( x - 1 )} d x + \int \frac{4}{3 ( x + 2 )} d x

\int \frac{3}{4 ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣

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

\int \frac{5}{12 ( x - 1 )} d x = \frac{5}{12} ln ∣ x - 1 ∣

\int \frac{4}{3 ( x + 2 )} d x = \frac{4}{3} ln ∣ x + 2 ∣

= - \frac{3}{4} ln ∣ x + 1 ∣ - (- \frac{3}{2 ( x + 1 )}) + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣

簡素化

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣

定数を解答に追加する

= - \frac{3}{4} ln ∣ x + 1 ∣ + \frac{3}{2 ( x + 1 )} + \frac{5}{12} ln ∣ x - 1 ∣ + \frac{4}{3} ln ∣ x + 2 ∣ + C