integral of 1/((x+2)(x+7)(2x-5)(x+1)^2)

解

\int \frac{1}{( x + 2 ) ( x + 7 ) ( 2 x - 5 ) ( x + 1 ) ^{2}} d x

- \frac{1}{45} ln ∣ x + 2 ∣ + \frac{1}{3420} ln ∣ x + 7 ∣ + \frac{8}{8379} ln ∣ 2 x - 5 ∣ + \frac{37}{1764} ln ∣ x + 1 ∣ + \frac{1}{42 ( x + 1 )} + C

解答ステップ

\int \frac{1}{( x + 2 ) ( x + 7 ) ( 2 x - 5 ) ( x + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{1}{( x + 2 ) ( x + 7 ) ( 2 x - 5 ) ( x + 1 ) ^{2}} : - \frac{1}{45 ( x + 2 )} + \frac{1}{3420 ( x + 7 )} + \frac{16}{8379 ( 2 x - 5 )} + \frac{37}{1764 ( x + 1 )} - \frac{1}{42 ( x + 1 ) ^{2}}

= \int - \frac{1}{45 ( x + 2 )} + \frac{1}{3420 ( x + 7 )} + \frac{16}{8379 ( 2 x - 5 )} + \frac{37}{1764 ( x + 1 )} - \frac{1}{42 ( 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{1}{45 ( x + 2 )} d x + \int \frac{1}{3420 ( x + 7 )} d x + \int \frac{16}{8379 ( 2 x - 5 )} d x + \int \frac{37}{1764 ( x + 1 )} d x - \int \frac{1}{42 ( x + 1 ) ^{2}} d x

\int \frac{1}{45 ( x + 2 )} d x = \frac{1}{45} ln ∣ x + 2 ∣

\int \frac{1}{3420 ( x + 7 )} d x = \frac{1}{3420} ln ∣ x + 7 ∣

\int \frac{16}{8379 ( 2 x - 5 )} d x = \frac{8}{8379} ln ∣ 2 x - 5 ∣

\int \frac{37}{1764 ( x + 1 )} d x = \frac{37}{1764} ln ∣ x + 1 ∣

\int \frac{1}{42 ( x + 1 ) ^{2}} d x = - \frac{1}{42 ( x + 1 )}

= - \frac{1}{45} ln ∣ x + 2 ∣ + \frac{1}{3420} ln ∣ x + 7 ∣ + \frac{8}{8379} ln ∣ 2 x - 5 ∣ + \frac{37}{1764} ln ∣ x + 1 ∣ - (- \frac{1}{42 ( x + 1 )})

簡素化

= - \frac{1}{45} ln ∣ x + 2 ∣ + \frac{1}{3420} ln ∣ x + 7 ∣ + \frac{8}{8379} ln ∣ 2 x - 5 ∣ + \frac{37}{1764} ln ∣ x + 1 ∣ + \frac{1}{42 ( x + 1 )}

定数を解答に追加する

= - \frac{1}{45} ln ∣ x + 2 ∣ + \frac{1}{3420} ln ∣ x + 7 ∣ + \frac{8}{8379} ln ∣ 2 x - 5 ∣ + \frac{37}{1764} ln ∣ x + 1 ∣ + \frac{1}{42 ( x + 1 )} + C