integral of (x/(x^2+6x+8))^2

解

\int (\frac{x}{x ^{2} + 6 x + 8})^{2} d x

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

解答ステップ

\int (\frac{x}{x ^{2} + 6 x + 8})^{2} d x

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int \frac{x ^{2}}{( x ^{2} + 6 x + 8 ) ^{2}} d x

以下の部分分数を得る:

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

= \int - \frac{2}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x + 4} + \frac{4}{( x + 4 ) ^{2}} 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 + 2} d x + \int \frac{1}{( x + 2 ) ^{2}} d x + \int \frac{2}{x + 4} d x + \int \frac{4}{( x + 4 ) ^{2}} d x

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

\int \frac{2}{x + 4} d x = 2 ln ∣ x + 4 ∣

\int \frac{4}{( x + 4 ) ^{2}} d x = - \frac{4}{x + 4}

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

定数を解答に追加する

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