integral of 1/(x(2x+3)^2)

解

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

\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ 2 x + 3 ∣ + \frac{1}{3 ( 2 x + 3 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ( 2 x + 3 ) ^{2}} : \frac{1}{9 x} - \frac{2}{9 ( 2 x + 3 )} - \frac{2}{3 ( 2 x + 3 ) ^{2}}

= \int \frac{1}{9 x} - \frac{2}{9 ( 2 x + 3 )} - \frac{2}{3 ( 2 x + 3 ) ^{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{1}{9 x} d x - \int \frac{2}{9 ( 2 x + 3 )} d x - \int \frac{2}{3 ( 2 x + 3 ) ^{2}} d x

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{2}{9 ( 2 x + 3 )} d x = \frac{1}{9} ln ∣ 2 x + 3 ∣

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

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ 2 x + 3 ∣ - (- \frac{1}{3 ( 2 x + 3 )})

簡素化

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ 2 x + 3 ∣ + \frac{1}{3 ( 2 x + 3 )}

定数を解答に追加する

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ 2 x + 3 ∣ + \frac{1}{3 ( 2 x + 3 )} + C