integral of (7x^2+13x)/(x^3+x^2-x-1)

解

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

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

解答ステップ

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

拡張

\frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} : \frac{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} + \frac{13 x}{x ^{3} + x ^{2} - x - 1}

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{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{13 x}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} d x = 7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

\int \frac{13 x}{x ^{3} + x ^{2} - x - 1} d x = 13 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

= 7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 13 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

簡素化

7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 13 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) : 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1}

= 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1}

定数を解答に追加する

= 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1} + C