integral of-1ln(x^2-1)

解答

\int - 1 ln (x^{2} - 1) d x

- x ln (x^{2} - 1) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x + C

求解步骤

\int - 1 \cdot ln (x^{2} - 1) d x

提出常数:

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

= - \int ln (x^{2} - 1) d x

使用分布积分法

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

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

= - (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

化简

= - x ln (x^{2} - 1) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x

解答补常数

= - x ln (x^{2} - 1) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x + C

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (0) = 2

\int sec (\frac{x}{5}) d x

in v erse l a pl a ce \frac{1}{s ( s ^{2} + 5 s + 6 )}

x \to 0 lim (x sin (\frac{1}{x ^{2}}))

d er i v a t i v e f (x) = \frac{x}{x - 1}