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

解

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

\frac{1}{2} ln x^{2} + 4 x + 3 + 3 ln ∣ x + 3 ∣ - 3 ln ∣ x + 1 ∣ + C

解答ステップ

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

拡張

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

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

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

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

簡素化

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

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

定数を解答に追加する

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