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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{( x - 1 ) ^{2}} d x = - \frac{1}{x - 1}

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

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

定数を解答に追加する

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