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

解

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

x + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 3 ∣ + C

解答ステップ

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

拡張

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

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

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

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

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

= x - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 3 ∣ + 2 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

簡素化

x - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 3 ∣ + 2 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : x + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 3 ∣

= x + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 3 ∣

定数を解答に追加する

= x + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{7}{2} ln ∣ x - 3 ∣ + C