integral of (17x+3)/((4x+1)(x-1))

解

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

\frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C

解答ステップ

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

以下の部分分数を得る:

\frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} : \frac{1}{4 x + 1} + \frac{4}{x - 1}

= \int \frac{1}{4 x + 1} + \frac{4}{x - 1} 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}{4 x + 1} d x + \int \frac{4}{x - 1} d x

\int \frac{1}{4 x + 1} d x = \frac{1}{4} ln ∣ 4 x + 1 ∣

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= \frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣

定数を解答に追加する

= \frac{1}{4} ln ∣ 4 x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C