integral of 1/((x+3)^2(x-5))

解

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

- \frac{1}{64} ln ∣ x + 3 ∣ + \frac{1}{8 ( x + 3 )} + \frac{1}{64} ln ∣ x - 5 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x + 3 ) ^{2} ( x - 5 )} : - \frac{1}{64 ( x + 3 )} - \frac{1}{8 ( x + 3 ) ^{2}} + \frac{1}{64 ( x - 5 )}

= \int - \frac{1}{64 ( x + 3 )} - \frac{1}{8 ( x + 3 ) ^{2}} + \frac{1}{64 ( x - 5 )} 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}{64 ( x + 3 )} d x - \int \frac{1}{8 ( x + 3 ) ^{2}} d x + \int \frac{1}{64 ( x - 5 )} d x

\int \frac{1}{64 ( x + 3 )} d x = \frac{1}{64} ln ∣ x + 3 ∣

\int \frac{1}{8 ( x + 3 ) ^{2}} d x = - \frac{1}{8 ( x + 3 )}

\int \frac{1}{64 ( x - 5 )} d x = \frac{1}{64} ln ∣ x - 5 ∣

= - \frac{1}{64} ln ∣ x + 3 ∣ - (- \frac{1}{8 ( x + 3 )}) + \frac{1}{64} ln ∣ x - 5 ∣

簡素化

= - \frac{1}{64} ln ∣ x + 3 ∣ + \frac{1}{8 ( x + 3 )} + \frac{1}{64} ln ∣ x - 5 ∣

定数を解答に追加する

= - \frac{1}{64} ln ∣ x + 3 ∣ + \frac{1}{8 ( x + 3 )} + \frac{1}{64} ln ∣ x - 5 ∣ + C