integral of (x^2-2x+11)/(x^2+3x-4)

解

\int \frac{x ^{2} - 2 x + 11}{x ^{2} + 3 x - 4} d x

x + 2 ln ∣ x - 1 ∣ - 7 ln ∣ x + 4 ∣ + C

解答ステップ

\int \frac{x ^{2} - 2 x + 11}{x ^{2} + 3 x - 4} d x

拡張

\frac{x ^{2} - 2 x + 11}{x ^{2} + 3 x - 4} : \frac{x ^{2}}{x ^{2} + 3 x - 4} - \frac{2 x}{x ^{2} + 3 x - 4} + \frac{11}{x ^{2} + 3 x - 4}

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 ^{2} + 3 x - 4} d x - \int \frac{2 x}{x ^{2} + 3 x - 4} d x + \int \frac{11}{x ^{2} + 3 x - 4} d x

\int \frac{x ^{2}}{x ^{2} + 3 x - 4} d x = x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣

\int \frac{2 x}{x ^{2} + 3 x - 4} d x = 2 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣)

\int \frac{11}{x ^{2} + 3 x - 4} d x = 11 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

= x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣ - 2 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣) + 11 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

簡素化

x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣ - 2 (\frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣) + 11 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣) : x + 2 ln ∣ x - 1 ∣ - 7 ln ∣ x + 4 ∣

= x + 2 ln ∣ x - 1 ∣ - 7 ln ∣ x + 4 ∣

定数を解答に追加する

= x + 2 ln ∣ x - 1 ∣ - 7 ln ∣ x + 4 ∣ + C