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受欢迎的微积分 >

integral of 1/(y^2+y-2)

解答

\int \frac{1}{y ^{2} + y - 2} d y

解答

- \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} y - ln - \frac{2}{3} + \frac{2}{3} y) + C

求解步骤

\int \frac{1}{y ^{2} + y - 2} d y

对

y^{2} + y - 2

配方:

(y + \frac{1}{2})^{2} - \frac{9}{4}

= \int \frac{1}{( y + \frac{1}{2} ) ^{2} - \frac{9}{4}} d y

使用换元积分法

= \int \frac{4}{4 u ^{2} - 9} d u

提出常数:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - 9} d u

使用换元积分法

= 4 \cdot \int \frac{1}{6 ( v ^{2} - 1 )} d v

提出常数:

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

因式分解

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

提出常数:

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

= 4 \cdot \frac{1}{6} (- \int \frac{1}{- v ^{2} + 1} d v)

使用积分常见定则:

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

= 4 \cdot \frac{1}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( y + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( y + \frac{1}{2} ) - 1}{2}))

化简

4 \cdot \frac{1}{6} (- (\frac{ln \frac{2}{3} ( y + \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( y + \frac{1}{2} ) - 1}{2})) : - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} y - ln - \frac{2}{3} + \frac{2}{3} y)

= - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} y - ln - \frac{2}{3} + \frac{2}{3} y)

解答补常数

= - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} y - ln - \frac{2}{3} + \frac{2}{3} y) + C

作图

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