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

解答

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

ln ∣ x ∣ - 3 arctan (x) + C

求解步骤

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - 3 arctan (x)

解答补常数

= ln ∣ x ∣ - 3 arctan (x) + C