integral of 1/(t^2-1)

解答

\int \frac{1}{t ^{2} - 1} d t

- \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣ + C

求解步骤

\int \frac{1}{t ^{2} - 1} d t

因式分解

t^{2} - 1 : - (- t^{2} + 1)

= \int \frac{1}{- ( - t ^{2} + 1 )} d t

提出常数:

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

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

使用积分常见定则:

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

= - (\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2})

化简

- (\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}) : - \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣

= - \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣

解答补常数

= - \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣ + C

\frac{d}{d x} (x^{2} + \frac{y ^{2}}{9}) = 1

l a pl a ce t r an s f or m t^{\frac{3}{2}}

a re a y = \frac{3}{2} x, y = 7 - x^{2}

\int \frac{t}{3 t ^{2} + 9} d t

y^{^{'}} = 8 y^{2}