integral of (x+5)/((x+3)(x-2)^2)

解

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

\frac{2}{25} ln ∣ x + 3 ∣ - \frac{2}{25} ln ∣ x - 2 ∣ - \frac{2}{x - 2} + \frac{3}{5 ( x - 2 )} + C

解答ステップ

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

拡張

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

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

\int \frac{x}{( x + 3 ) ( x - 2 ) ^{2}} d x = - \frac{1}{x - 2} - 3 (\frac{1}{25} ln ∣ x + 3 ∣ - \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{5 ( x - 2 )})

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

= - \frac{1}{x - 2} - 3 (\frac{1}{25} ln ∣ x + 3 ∣ - \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{5 ( x - 2 )}) + \frac{1}{5} ln ∣ x + 3 ∣ - \frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{x - 2}

簡素化

- \frac{1}{x - 2} - 3 (\frac{1}{25} ln ∣ x + 3 ∣ - \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{5 ( x - 2 )}) + \frac{1}{5} ln ∣ x + 3 ∣ - \frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{x - 2} : \frac{2}{25} ln ∣ x + 3 ∣ - \frac{2}{25} ln ∣ x - 2 ∣ - \frac{2}{x - 2} + \frac{3}{5 ( x - 2 )}

= \frac{2}{25} ln ∣ x + 3 ∣ - \frac{2}{25} ln ∣ x - 2 ∣ - \frac{2}{x - 2} + \frac{3}{5 ( x - 2 )}

定数を解答に追加する

= \frac{2}{25} ln ∣ x + 3 ∣ - \frac{2}{25} ln ∣ x - 2 ∣ - \frac{2}{x - 2} + \frac{3}{5 ( x - 2 )} + C