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

解

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

\frac{2}{x + 1} + ln ∣ x - 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

簡素化

= \frac{2}{x + 1} + ln ∣ x - 1 ∣

定数を解答に追加する

= \frac{2}{x + 1} + ln ∣ x - 1 ∣ + C