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

解

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

x + \frac{1}{2} (x + 1)^{2} - 6 ln ∣ x + 1 ∣ + 1 + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ + 3 (x + 1 - ln ∣ x + 1 ∣) - 4 ln ∣ x + 1 ∣ : x + \frac{1}{2} (x + 1)^{2} - 6 ln ∣ x + 1 ∣ + 1

= x + \frac{1}{2} (x + 1)^{2} - 6 ln ∣ x + 1 ∣ + 1

定数を解答に追加する

= x + \frac{1}{2} (x + 1)^{2} - 6 ln ∣ x + 1 ∣ + 1 + C