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

解

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

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

解答ステップ

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

拡張

\frac{x ^{3} + 4}{x ( x + 1 )} : \frac{x ^{3}}{x ( x + 1 )} + \frac{4}{x ( x + 1 )}

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

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

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

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

簡素化

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

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

定数を解答に追加する

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