integral of (5x^3+20x^2-4x+4)/(x^2+4x)

解

\int \frac{5 x ^{3} + 20 x ^{2} - 4 x + 4}{x ^{2} + 4 x} d x

- 20 x + ln ∣ x ∣ + \frac{5}{2} (x + 4)^{2} - 5 ln ∣ x + 4 ∣ - 80 + C

解答ステップ

\int \frac{5 x ^{3} + 20 x ^{2} - 4 x + 4}{x ^{2} + 4 x} d x

拡張

\frac{5 x ^{3} + 20 x ^{2} - 4 x + 4}{x ^{2} + 4 x} : \frac{5 x ^{3}}{x ^{2} + 4 x} + \frac{20 x ^{2}}{x ^{2} + 4 x} - \frac{4 x}{x ^{2} + 4 x} + \frac{4}{x ^{2} + 4 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{5 x ^{3}}{x ^{2} + 4 x} d x + \int \frac{20 x ^{2}}{x ^{2} + 4 x} d x - \int \frac{4 x}{x ^{2} + 4 x} d x + \int \frac{4}{x ^{2} + 4 x} d x

\int \frac{5 x ^{3}}{x ^{2} + 4 x} d x = 5 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣)

\int \frac{20 x ^{2}}{x ^{2} + 4 x} d x = 20 (x + 4 - 4 ln ∣ x + 4 ∣)

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

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

= 5 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣) + 20 (x + 4 - 4 ln ∣ x + 4 ∣) - 4 ln ∣ x + 4 ∣ + ln ∣ x ∣ - ln ∣ x + 4 ∣

簡素化

5 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣) + 20 (x + 4 - 4 ln ∣ x + 4 ∣) - 4 ln ∣ x + 4 ∣ + ln ∣ x ∣ - ln ∣ x + 4 ∣ : - 20 x + ln ∣ x ∣ + \frac{5}{2} (x + 4)^{2} - 5 ln ∣ x + 4 ∣ - 80

= - 20 x + ln ∣ x ∣ + \frac{5}{2} (x + 4)^{2} - 5 ln ∣ x + 4 ∣ - 80

定数を解答に追加する

= - 20 x + ln ∣ x ∣ + \frac{5}{2} (x + 4)^{2} - 5 ln ∣ x + 4 ∣ - 80 + C