integral of x/((x-6)(x+2)^2)

解

\int \frac{x}{( x - 6 ) ( x + 2 ) ^{2}} d x

\frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} + C

解答ステップ

\int \frac{x}{( x - 6 ) ( x + 2 ) ^{2}} d x

以下の部分分数を得る:

\frac{x}{( x - 6 ) ( x + 2 ) ^{2}} : \frac{3}{32 ( x - 6 )} - \frac{3}{32 ( x + 2 )} + \frac{1}{4 ( x + 2 ) ^{2}}

= \int \frac{3}{32 ( x - 6 )} - \frac{3}{32 ( x + 2 )} + \frac{1}{4 ( 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}{32 ( x - 6 )} d x - \int \frac{3}{32 ( x + 2 )} d x + \int \frac{1}{4 ( x + 2 ) ^{2}} d x

\int \frac{3}{32 ( x - 6 )} d x = \frac{3}{32} ln ∣ x - 6 ∣

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

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

= \frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )}

定数を解答に追加する

= \frac{3}{32} ln ∣ x - 6 ∣ - \frac{3}{32} ln ∣ x + 2 ∣ - \frac{1}{4 ( x + 2 )} + C