integral of ((2x-1))/(((x+1)(x^2+4x+4)))

解

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

- 3 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ - \frac{5}{x + 2} + C

解答ステップ

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

以下の部分分数を得る:

\frac{2 x - 1}{( x + 1 ) ( x ^{2} + 4 x + 4 )} : - \frac{3}{x + 1} + \frac{3}{x + 2} + \frac{5}{( x + 2 ) ^{2}}

= \int - \frac{3}{x + 1} + \frac{3}{x + 2} + \frac{5}{( x + 2 ) ^{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{3}{x + 1} d x + \int \frac{3}{x + 2} d x + \int \frac{5}{( x + 2 ) ^{2}} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{5}{( x + 2 ) ^{2}} d x = - \frac{5}{x + 2}

= - 3 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ - \frac{5}{x + 2}

定数を解答に追加する

= - 3 ln ∣ x + 1 ∣ + 3 ln ∣ x + 2 ∣ - \frac{5}{x + 2} + C