integral of (10)/(x^2-1)

解答

\int \frac{10}{x ^{2} - 1} d x

- 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

求解步骤

\int \frac{10}{x ^{2} - 1} d x

提出常数:

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

= 10 \cdot \int \frac{1}{x ^{2} - 1} d x

因式分解

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

= 10 \cdot \int \frac{1}{- ( - x ^{2} + 1 )} d x

提出常数:

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

= 10 (- \int \frac{1}{- x ^{2} + 1} d x)

使用积分常见定则:

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

= 10 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

化简

10 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

解答补常数

= - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

a re a y = x^{\frac{1}{2}}, y = x - 2, [0, 4]

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

l a pl a ce t r an s f or m cos^{3} (t)

\frac{d y}{d x} = \frac{6 y e ^{3 x}}{2 + e ^{3 x}}

\int_{- \infty}^{- 1} e^{- 28 t} d t