integral of (29x^4+29x^3+2x+1)/(x^2+x)

解

\int \frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} d x

29 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{29}{3} (x + 1)^{3} + \frac{29}{2} (x + 1)^{2} - \frac{87}{2} (x + 1)^{2} + 29 + C

解答ステップ

\int \frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} d x

拡張

\frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} : \frac{29 x ^{4}}{x ^{2} + x} + \frac{29 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{29 x ^{4}}{x ^{2} + x} d x + \int \frac{29 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{29 x ^{4}}{x ^{2} + x} d x = 29 (\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣)

\int \frac{29 x ^{3}}{x ^{2} + x} d x = 29 (\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 ∣

= 29 (\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣) + 29 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣

簡素化

29 (\frac{( x + 1 ) ^{3}}{3} - \frac{3 ( x + 1 ) ^{2}}{2} + 3 (x + 1) - ln ∣ x + 1 ∣) + 29 (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) + 2 ln ∣ x + 1 ∣ + ln ∣ x ∣ - ln ∣ x + 1 ∣ : 29 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{29}{3} (x + 1)^{3} + \frac{29}{2} (x + 1)^{2} - \frac{87}{2} (x + 1)^{2} + 29

= 29 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{29}{3} (x + 1)^{3} + \frac{29}{2} (x + 1)^{2} - \frac{87}{2} (x + 1)^{2} + 29

定数を解答に追加する

= 29 x + ln ∣ x + 1 ∣ + ln ∣ x ∣ + \frac{29}{3} (x + 1)^{3} + \frac{29}{2} (x + 1)^{2} - \frac{87}{2} (x + 1)^{2} + 29 + C