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

解

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

- \frac{1}{25} ln ∣ x + 2 ∣ + \frac{1}{5 ( x + 2 )} + \frac{1}{25} ln ∣ x - 3 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x + 2 ) ^{2} ( x - 3 )} : - \frac{1}{25 ( x + 2 )} - \frac{1}{5 ( x + 2 ) ^{2}} + \frac{1}{25 ( x - 3 )}

= \int - \frac{1}{25 ( x + 2 )} - \frac{1}{5 ( x + 2 ) ^{2}} + \frac{1}{25 ( x - 3 )} 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 + 2 )} d x - \int \frac{1}{5 ( x + 2 ) ^{2}} d x + \int \frac{1}{25 ( x - 3 )} d x

\int \frac{1}{25 ( x + 2 )} d x = \frac{1}{25} ln ∣ x + 2 ∣

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

\int \frac{1}{25 ( x - 3 )} d x = \frac{1}{25} ln ∣ x - 3 ∣

= - \frac{1}{25} ln ∣ x + 2 ∣ - (- \frac{1}{5 ( x + 2 )}) + \frac{1}{25} ln ∣ x - 3 ∣

簡素化

= - \frac{1}{25} ln ∣ x + 2 ∣ + \frac{1}{5 ( x + 2 )} + \frac{1}{25} ln ∣ x - 3 ∣

定数を解答に追加する

= - \frac{1}{25} ln ∣ x + 2 ∣ + \frac{1}{5 ( x + 2 )} + \frac{1}{25} ln ∣ x - 3 ∣ + C