integral of 8ln(x^2-1)

解答

\int 8 ln (x^{2} - 1) d x

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

求解步骤

\int 8 ln (x^{2} - 1) d x

提出常数:

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

= 8 \cdot \int ln (x^{2} - 1) d x

使用分布积分法

= 8 (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

= 8 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

化简

= 8 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

解答补常数

= 8 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

f (x) = \frac{1}{x ln ( x )}

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

d er i v a t i v e tan (3 t)

\int 2 t 2 t - 3 d t

x^{^{'}} (t) = (\frac{1}{3} x) (1 - \frac{x}{3}), x (0) = 6