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

解

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

- \frac{7}{x + 2} - \frac{5}{( x + 2 ) ^{2}} - 3 ln (\frac{1}{2} ∣ x ∣) + 3 ln \frac{x}{2} + 1 + 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 = 6 (\frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{1}{2} ln (\frac{1}{2} ∣ x ∣) - \frac{1}{2} ln \frac{x}{2} + 1)

= - \frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}} - 6 (\frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{1}{2} ln (\frac{1}{2} ∣ x ∣) - \frac{1}{2} ln \frac{x}{2} + 1)

簡素化

- \frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}} - 6 (\frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{1}{2} ln (\frac{1}{2} ∣ x ∣) - \frac{1}{2} ln \frac{x}{2} + 1) : - \frac{7}{x + 2} - \frac{5}{( x + 2 ) ^{2}} - 3 ln (\frac{1}{2} ∣ x ∣) + 3 ln \frac{x}{2} + 1

= - \frac{7}{x + 2} - \frac{5}{( x + 2 ) ^{2}} - 3 ln (\frac{1}{2} ∣ x ∣) + 3 ln \frac{x}{2} + 1

定数を解答に追加する

= - \frac{7}{x + 2} - \frac{5}{( x + 2 ) ^{2}} - 3 ln (\frac{1}{2} ∣ x ∣) + 3 ln \frac{x}{2} + 1 + C