integral of (5x^4+5x^3+2x+1)/(x^2+x)

解

\int \frac{5 x ^{4} + 5 x ^{3} + 2 x + 1}{x ^{2} + x} d x

5 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{5}{3} (x + 1)^{3} + \frac{5}{2} (x + 1)^{2} - \frac{15}{2} (x + 1)^{2} + 5 + C

解答ステップ

\int \frac{5 x ^{4} + 5 x ^{3} + 2 x + 1}{x ^{2} + x} d x

拡張

\frac{5 x ^{4} + 5 x ^{3} + 2 x + 1}{x ^{2} + x} : \frac{5 x ^{4}}{x ^{2} + x} + \frac{5 x ^{3}}{x ^{2} + x} + \frac{2 x}{x ^{2} + x} + \frac{1}{x ^{2} + 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 ^{4}}{x ^{2} + x} d x + \int \frac{5 x ^{3}}{x ^{2} + x} d x + \int \frac{2 x}{x ^{2} + x} d x + \int \frac{1}{x ^{2} + x} d x

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

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

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

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

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

簡素化

5 (\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣) + 5 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣ : 5 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{5}{3} (x + 1)^{3} + \frac{5}{2} (x + 1)^{2} - \frac{15}{2} (x + 1)^{2} + 5

= 5 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{5}{3} (x + 1)^{3} + \frac{5}{2} (x + 1)^{2} - \frac{15}{2} (x + 1)^{2} + 5

定数を解答に追加する

= 5 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{5}{3} (x + 1)^{3} + \frac{5}{2} (x + 1)^{2} - \frac{15}{2} (x + 1)^{2} + 5 + C