integral of (2x^2+16)/(x(x-2)^3)

解

\int \frac{2 x ^{2} + 16}{x ( x - 2 ) ^{3}} d x

\frac{2}{x - 2} - \frac{6}{( x - 2 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{2 x ^{2} + 16}{x ( x - 2 ) ^{3}} d x

拡張

\frac{2 x ^{2} + 16}{x ( x - 2 ) ^{3}} : \frac{2 x}{( x - 2 ) ^{3}} + \frac{16}{x ( x - 2 ) ^{3}}

= \int \frac{2 x}{( x - 2 ) ^{3}} + \frac{16}{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{2 x}{( x - 2 ) ^{3}} d x + \int \frac{16}{x ( x - 2 ) ^{3}} d x

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

\int \frac{16}{x ( x - 2 ) ^{3}} d x = 16 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 ( x - 2 )} - \frac{1}{4 ( x - 2 ) ^{2}})

= 2 (- \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}}) + 16 (- \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

2 (- \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}}) + 16 (- \frac{1}{8} ln ∣ x ∣ + \frac{1}{8} ln ∣ x - 2 ∣ + \frac{1}{4 ( x - 2 )} - \frac{1}{4 ( x - 2 ) ^{2}}) : \frac{2}{x - 2} - \frac{6}{( x - 2 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣

= \frac{2}{x - 2} - \frac{6}{( x - 2 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{2}{x - 2} - \frac{6}{( x - 2 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x - 2 ∣ + C