integral of (x^2-64)/(x(x-1)^2)

解

\int \frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} d x

65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣ + C

解答ステップ

\int \frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} d x

拡張

\frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} : \frac{x ^{2}}{x ( x - 1 ) ^{2}} - \frac{64}{x ( x - 1 ) ^{2}}

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 ^{2}}{x ( x - 1 ) ^{2}} d x - \int \frac{64}{x ( x - 1 ) ^{2}} d x

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

\int \frac{64}{x ( x - 1 ) ^{2}} d x = 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1})

= ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1})

簡素化

ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1}) : 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣

= 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣

定数を解答に追加する

= 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣ + C