integral of (x^2+24)/(x(x-2)^3)

解

\int \frac{x ^{2} + 24}{x ( x - 2 ) ^{3}} d x

\frac{5}{x - 2} - \frac{7}{( x - 2 ) ^{2}} - 3 ln ∣ x ∣ + 3 ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{x ^{2} + 24}{x ( x - 2 ) ^{3}} d x

拡張

\frac{x ^{2} + 24}{x ( x - 2 ) ^{3}} : \frac{x}{( x - 2 ) ^{3}} + \frac{24}{x ( x - 2 ) ^{3}}

= \int \frac{x}{( x - 2 ) ^{3}} + \frac{24}{x ( x - 2 ) ^{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{x}{( x - 2 ) ^{3}} d x + \int \frac{24}{x ( x - 2 ) ^{3}} d x

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

\int \frac{24}{x ( x - 2 ) ^{3}} d x = 24 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 ( x - 2 )} - \frac{1}{4 ( x - 2 ) ^{2}})

= - \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}} + 24 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 ( x - 2 )} - \frac{1}{4 ( x - 2 ) ^{2}})

簡素化

- \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}} + 24 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 ( x - 2 )} - \frac{1}{4 ( x - 2 ) ^{2}}) : \frac{5}{x - 2} - \frac{7}{( x - 2 ) ^{2}} - 3 ln ∣ x ∣ + 3 ln ∣ x - 2 ∣

= \frac{5}{x - 2} - \frac{7}{( x - 2 ) ^{2}} - 3 ln ∣ x ∣ + 3 ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{5}{x - 2} - \frac{7}{( x - 2 ) ^{2}} - 3 ln ∣ x ∣ + 3 ln ∣ x - 2 ∣ + C