integral of (x^2)/((2x+3)(4x^2-1))

解

\int \frac{x ^{2}}{( 2 x + 3 ) ( 4 x ^{2} - 1 )} d x

\frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣ + C

解答ステップ

\int \frac{x ^{2}}{( 2 x + 3 ) ( 4 x ^{2} - 1 )} d x

以下の部分分数を得る:

\frac{x ^{2}}{( 2 x + 3 ) ( 4 x ^{2} - 1 )} : \frac{9}{32 ( 2 x + 3 )} - \frac{1}{16 ( 2 x + 1 )} + \frac{1}{32 ( 2 x - 1 )}

= \int \frac{9}{32 ( 2 x + 3 )} - \frac{1}{16 ( 2 x + 1 )} + \frac{1}{32 ( 2 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{9}{32 ( 2 x + 3 )} d x - \int \frac{1}{16 ( 2 x + 1 )} d x + \int \frac{1}{32 ( 2 x - 1 )} d x

\int \frac{9}{32 ( 2 x + 3 )} d x = \frac{9}{64} ln ∣ 2 x + 3 ∣

\int \frac{1}{16 ( 2 x + 1 )} d x = \frac{1}{32} ln ∣ 2 x + 1 ∣

\int \frac{1}{32 ( 2 x - 1 )} d x = \frac{1}{64} ln ∣ 2 x - 1 ∣

= \frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣

定数を解答に追加する

= \frac{9}{64} ln ∣ 2 x + 3 ∣ - \frac{1}{32} ln ∣ 2 x + 1 ∣ + \frac{1}{64} ln ∣ 2 x - 1 ∣ + C