integral of 4/((x^2+5x-14))

解

\int \frac{4}{( x ^{2} + 5 x - 14 )} d x

4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣) + C

解答ステップ

\int \frac{4}{x ^{2} + 5 x - 14} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{x ^{2} + 5 x - 14} d x

以下の部分分数を得る:

\frac{1}{x ^{2} + 5 x - 14} : \frac{1}{9 ( x - 2 )} - \frac{1}{9 ( x + 7 )}

= 4 \cdot \int \frac{1}{9 ( x - 2 )} - \frac{1}{9 ( x + 7 )} 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

= 4 (\int \frac{1}{9 ( x - 2 )} d x - \int \frac{1}{9 ( x + 7 )} d x)

\int \frac{1}{9 ( x - 2 )} d x = \frac{1}{9} ln ∣ x - 2 ∣

\int \frac{1}{9 ( x + 7 )} d x = \frac{1}{9} ln ∣ x + 7 ∣

= 4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣)

定数を解答に追加する

= 4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣) + C