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

解

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

\frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x + 4 ) ( x - 1 ) ^{2}} : \frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{2}}

= \int \frac{1}{25 ( x + 4 )} - \frac{1}{25 ( x - 1 )} + \frac{1}{5 ( x - 1 ) ^{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{1}{25 ( x + 4 )} d x - \int \frac{1}{25 ( x - 1 )} d x + \int \frac{1}{5 ( x - 1 ) ^{2}} d x

\int \frac{1}{25 ( x + 4 )} d x = \frac{1}{25} ln ∣ x + 4 ∣

\int \frac{1}{25 ( x - 1 )} d x = \frac{1}{25} ln ∣ x - 1 ∣

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

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )}

定数を解答に追加する

= \frac{1}{25} ln ∣ x + 4 ∣ - \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{5 ( x - 1 )} + C