integral of-4ln(x^2-1)

解答

\int - 4 ln (x^{2} - 1) d x

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

求解步骤

\int - 4 ln (x^{2} - 1) d x

提出常数:

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

= - 4 \cdot \int ln (x^{2} - 1) d x

使用分布积分法

= - 4 (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

= - 4 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

化简

= - 4 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

解答补常数

= - 4 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

n = 1 \sum \infty n (1)^{n}

t an g e n t 7 x^{2} + 8 x

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

\int (8 x^{3} + \frac{1}{2 x ^{2}}) d x

in v erse l a pl a ce \frac{( 3 s - 4 )}{s ^{2} + 9}