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

解

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

3 ln ∣ x ∣ - \frac{64}{27} ln ∣ 2 x - 1 ∣ - \frac{14}{9 ( 2 x - 1 )} - \frac{17}{27} ln ∣ x + 1 ∣ + \frac{2}{9 ( x + 1 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{x + 3}{x ( 2 x - 1 ) ^{2} ( x + 1 ) ^{2}} : \frac{3}{x} - \frac{128}{27 ( 2 x - 1 )} + \frac{28}{9 ( 2 x - 1 ) ^{2}} - \frac{17}{27 ( x + 1 )} - \frac{2}{9 ( x + 1 ) ^{2}}

= \int \frac{3}{x} - \frac{128}{27 ( 2 x - 1 )} + \frac{28}{9 ( 2 x - 1 ) ^{2}} - \frac{17}{27 ( x + 1 )} - \frac{2}{9 ( x + 1 ) ^{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} d x - \int \frac{128}{27 ( 2 x - 1 )} d x + \int \frac{28}{9 ( 2 x - 1 ) ^{2}} d x - \int \frac{17}{27 ( x + 1 )} d x - \int \frac{2}{9 ( x + 1 ) ^{2}} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{128}{27 ( 2 x - 1 )} d x = \frac{64}{27} ln ∣ 2 x - 1 ∣

\int \frac{28}{9 ( 2 x - 1 ) ^{2}} d x = - \frac{14}{9 ( 2 x - 1 )}

\int \frac{17}{27 ( x + 1 )} d x = \frac{17}{27} ln ∣ x + 1 ∣

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

= 3 ln ∣ x ∣ - \frac{64}{27} ln ∣ 2 x - 1 ∣ - \frac{14}{9 ( 2 x - 1 )} - \frac{17}{27} ln ∣ x + 1 ∣ - (- \frac{2}{9 ( x + 1 )})

簡素化

= 3 ln ∣ x ∣ - \frac{64}{27} ln ∣ 2 x - 1 ∣ - \frac{14}{9 ( 2 x - 1 )} - \frac{17}{27} ln ∣ x + 1 ∣ + \frac{2}{9 ( x + 1 )}

定数を解答に追加する

= 3 ln ∣ x ∣ - \frac{64}{27} ln ∣ 2 x - 1 ∣ - \frac{14}{9 ( 2 x - 1 )} - \frac{17}{27} ln ∣ x + 1 ∣ + \frac{2}{9 ( x + 1 )} + C