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

解

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

2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + C

解答ステップ

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

拡張

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

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

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

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

簡素化

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

= 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2}

定数を解答に追加する

= 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + C