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

解

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

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

解答ステップ

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

拡張

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

\int \frac{x ^{2}}{x + 3} d x = \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣

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

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

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ + 4 (x + 3 - 3 ln ∣ x + 3 ∣) + ln ∣ x + 3 ∣

簡素化

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ + 4 (x + 3 - 3 ln ∣ x + 3 ∣) + ln ∣ x + 3 ∣ : - 2 x + \frac{1}{2} (x + 3)^{2} - 2 ln ∣ x + 3 ∣ - 6

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

定数を解答に追加する

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