integral of (7x+x^2)/((x-4)(2x+3)(2x+1))

解

\int \frac{7 x + x ^{2}}{( x - 4 ) ( 2 x + 3 ) ( 2 x + 1 )} d x

\frac{4}{9} ln ∣ x - 4 ∣ - \frac{3}{8} ln ∣ 2 x + 3 ∣ + \frac{13}{72} ln ∣ 2 x + 1 ∣ + C

解答ステップ

\int \frac{7 x + x ^{2}}{( x - 4 ) ( 2 x + 3 ) ( 2 x + 1 )} d x

以下の部分分数を得る:

\frac{7 x + x ^{2}}{( x - 4 ) ( 2 x + 3 ) ( 2 x + 1 )} : \frac{4}{9 ( x - 4 )} - \frac{3}{4 ( 2 x + 3 )} + \frac{13}{36 ( 2 x + 1 )}

= \int \frac{4}{9 ( x - 4 )} - \frac{3}{4 ( 2 x + 3 )} + \frac{13}{36 ( 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{4}{9 ( x - 4 )} d x - \int \frac{3}{4 ( 2 x + 3 )} d x + \int \frac{13}{36 ( 2 x + 1 )} d x

\int \frac{4}{9 ( x - 4 )} d x = \frac{4}{9} ln ∣ x - 4 ∣

\int \frac{3}{4 ( 2 x + 3 )} d x = \frac{3}{8} ln ∣ 2 x + 3 ∣

\int \frac{13}{36 ( 2 x + 1 )} d x = \frac{13}{72} ln ∣ 2 x + 1 ∣

= \frac{4}{9} ln ∣ x - 4 ∣ - \frac{3}{8} ln ∣ 2 x + 3 ∣ + \frac{13}{72} ln ∣ 2 x + 1 ∣

定数を解答に追加する

= \frac{4}{9} ln ∣ x - 4 ∣ - \frac{3}{8} ln ∣ 2 x + 3 ∣ + \frac{13}{72} ln ∣ 2 x + 1 ∣ + C