integral of (4x^2+13x-9)/(x^3+2x^2-3x)

解

\int \frac{4 x ^{2} + 13 x - 9}{x ^{3} + 2 x ^{2} - 3 x} d x

2 ln x^{2} + 2 x - 3 + ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1 - \frac{13}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) + C

解答ステップ

\int \frac{4 x ^{2} + 13 x - 9}{x ^{3} + 2 x ^{2} - 3 x} d x

拡張

\frac{4 x ^{2} + 13 x - 9}{x ^{3} + 2 x ^{2} - 3 x} : \frac{4 x ^{2}}{x ^{3} + 2 x ^{2} - 3 x} + \frac{13 x}{x ^{3} + 2 x ^{2} - 3 x} - \frac{9}{x ^{3} + 2 x ^{2} - 3 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{4 x ^{2}}{x ^{3} + 2 x ^{2} - 3 x} d x + \int \frac{13 x}{x ^{3} + 2 x ^{2} - 3 x} d x - \int \frac{9}{x ^{3} + 2 x ^{2} - 3 x} d x

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

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

\int \frac{9}{x ^{3} + 2 x ^{2} - 3 x} d x = 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

= 2 ln x^{2} + 2 x - 3 + ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1 - \frac{13}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

定数を解答に追加する

= 2 ln x^{2} + 2 x - 3 + ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1 - \frac{13}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) + C