integral of (7x^3+8x)/(x-1)

解

\int \frac{7 x ^{3} + 8 x}{x - 1} d x

\frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣ + C

解答ステップ

\int \frac{7 x ^{3} + 8 x}{x - 1} d x

長除法

\frac{7 x ^{3} + 8 x}{x - 1} : 7 x^{2} + 7 x + 15 + \frac{15}{x - 1}

= \int 7 x^{2} + 7 x + 15 + \frac{15}{x - 1} d 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 7 x^{2} d x + \int 7 x d x + \int 15 d x + \int \frac{15}{x - 1} d x

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 15 d x = 15 x

\int \frac{15}{x - 1} d x = 15 ln ∣ x - 1 ∣

= \frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣

定数を解答に追加する

= \frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣ + C