integral of 1/(2x^2+3x-2)

解

\int \frac{1}{2 x ^{2} + 3 x - 2} d x

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

解答ステップ

\int \frac{1}{2 x ^{2} + 3 x - 2} d x

以下の部分分数を得る:

\frac{1}{2 x ^{2} + 3 x - 2} : \frac{2}{5 ( 2 x - 1 )} - \frac{1}{5 ( x + 2 )}

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

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

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

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

定数を解答に追加する

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