integral of (x^2-5x+16)/((2x+1)(x-2)^2)

解

\int \frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} d x

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

解答ステップ

\int \frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} d x

以下の部分分数を得る:

\frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} : \frac{3}{2 x + 1} - \frac{1}{x - 2} + \frac{2}{( x - 2 ) ^{2}}

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

\int \frac{3}{2 x + 1} d x = \frac{3}{2} ln ∣ 2 x + 1 ∣

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

\int \frac{2}{( x - 2 ) ^{2}} d x = - \frac{2}{x - 2}

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

定数を解答に追加する

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