integral of (ax+b)/(cx+

解

\int \frac{a x + b}{c x + d} d x

a \frac{1}{c ^{2}} (c x + d - d ln ∣ c x + d ∣) + b \frac{1}{c} ln ∣ c x + d ∣ + C

解答ステップ

\int \frac{a x + b}{c x + d} d x

拡張

\frac{a x + b}{c x + d} : \frac{a x}{c x + d} + \frac{b}{c x + d}

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{a x}{c x + d} d x + \int \frac{b}{c x + d} d x

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

\int \frac{b}{c x + d} d x = b \frac{1}{c} ln ∣ c x + d ∣

= a \frac{1}{c ^{2}} (c x + d - d ln ∣ c x + d ∣) + b \frac{1}{c} ln ∣ c x + d ∣

定数を解答に追加する

= a \frac{1}{c ^{2}} (c x + d - d ln ∣ c x + d ∣) + b \frac{1}{c} ln ∣ c x + d ∣ + C