integral of (x^2+1)/(x^2-5x+6)

解答

\int \frac{x ^{2} + 1}{x ^{2} - 5 x + 6} d x

10 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) - 5 (- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣) + x + C

求解步骤

\int \frac{x ^{2} + 1}{x ^{2} - 5 x + 6} d x

\frac{x ^{2} + 1}{x ^{2} - 5 x + 6} = \frac{5 x - 5}{x ^{2} - 5 x + 6} + 1

= \int \frac{5 x - 5}{x ^{2} - 5 x + 6} + 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{5 x - 5}{x ^{2} - 5 x + 6} d x + \int 1 d x

\int \frac{5 x - 5}{x ^{2} - 5 x + 6} d x = 10 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) - 5 (- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣)

\int 1 d x = x

= 10 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) - 5 (- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣) + x

解答补常数

= 10 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) - 5 (- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣) + x + C

\frac{\partial}{\partial x} (5 e^{2 x})

d er i v a t i v e y = \frac{x ^{4}}{4} + \frac{2}{9} x^{3} - x^{2} + 9 x - 5

ma c l a u r in \frac{( 2 x )}{1 + x ^{4}}

d er i v a t i v e 7 - 4 x

\frac{d}{d x} (e^{4 x s i n (2 x)})