integral of (2x^2-8)/(x(x+2)^3)

解

\int \frac{2 x ^{2} - 8}{x ( x + 2 ) ^{3}} d x

- \frac{4}{x + 2} - ln (\frac{1}{2} ∣ x ∣) + ln \frac{x}{2} + 1 + C

解答ステップ

\int \frac{2 x ^{2} - 8}{x ( x + 2 ) ^{3}} d x

拡張

\frac{2 x ^{2} - 8}{x ( x + 2 ) ^{3}} : \frac{2 x ^{2}}{x ( x + 2 ) ^{3}} - \frac{8}{x ( x + 2 ) ^{3}}

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 ^{2}}{x ( x + 2 ) ^{3}} d x - \int \frac{8}{x ( x + 2 ) ^{3}} d x

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

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

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

簡素化

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

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

定数を解答に追加する

= - \frac{4}{x + 2} - ln (\frac{1}{2} ∣ x ∣) + ln \frac{x}{2} + 1 + C