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

解

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

\frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{x ( 2 x + 5 ) ^{2}} : \frac{1}{25 x} - \frac{2}{25 ( 2 x + 5 )} - \frac{2}{5 ( 2 x + 5 ) ^{2}}

= \int \frac{1}{25 x} - \frac{2}{25 ( 2 x + 5 )} - \frac{2}{5 ( 2 x + 5 ) ^{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}{25 x} d x - \int \frac{2}{25 ( 2 x + 5 )} d x - \int \frac{2}{5 ( 2 x + 5 ) ^{2}} d x

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{2}{25 ( 2 x + 5 )} d x = \frac{1}{25} ln ∣ 2 x + 5 ∣

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

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ - (- \frac{1}{5 ( 2 x + 5 )})

簡素化

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )}

定数を解答に追加する

= \frac{1}{25} ln ∣ x ∣ - \frac{1}{25} ln ∣ 2 x + 5 ∣ + \frac{1}{5 ( 2 x + 5 )} + C