integral of 3/(x^2+4x-5)

解

\int \frac{3}{x ^{2} + 4 x - 5} d x

3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) + C

解答ステップ

\int \frac{3}{x ^{2} + 4 x - 5} d x

定数を除く:

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

= 3 \cdot \int \frac{1}{x ^{2} + 4 x - 5} d x

以下の部分分数を得る:

\frac{1}{x ^{2} + 4 x - 5} : \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 5 )}

= 3 \cdot \int \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( 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

= 3 (\int \frac{1}{6 ( x - 1 )} d x - \int \frac{1}{6 ( x + 5 )} d x)

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

\int \frac{1}{6 ( x + 5 )} d x = \frac{1}{6} ln ∣ x + 5 ∣

= 3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

定数を解答に追加する

= 3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) + C