integral of 1/((x^2+10x+24))

解

\int \frac{1}{( x ^{2} + 10 x + 24 )} d x

\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣ + C

解答ステップ

\int \frac{1}{x ^{2} + 10 x + 24} d x

以下の部分分数を得る:

\frac{1}{x ^{2} + 10 x + 24} : \frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )}

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

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

= \frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣

定数を解答に追加する

= \frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣ + C