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受欢迎的微积分 >

integral of 7/(2x^2-5x-3)

解答

\int \frac{7}{2 x ^{2} - 5 x - 3} d x

解答

- ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x + C

求解步骤

\int \frac{7}{2 x ^{2} - 5 x - 3} d x

提出常数:

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

= 7 \cdot \int \frac{1}{2 x ^{2} - 5 x - 3} d x

对

2 x^{2} - 5 x - 3

配方:

2 (x - \frac{5}{4})^{2} - \frac{49}{8}

= 7 \cdot \int \frac{1}{2 ( x - \frac{5}{4} ) ^{2} - \frac{49}{8}} d x

使用换元积分法

= 7 \cdot \int \frac{8}{16 u ^{2} - 49} d u

提出常数:

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

= 7 \cdot 8 \cdot \int \frac{1}{16 u ^{2} - 49} d u

使用换元积分法

= 7 \cdot 8 \cdot \int \frac{1}{28 ( v ^{2} - 1 )} d v

提出常数:

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

= 7 \cdot 8 \cdot \frac{1}{28} \cdot \int \frac{1}{v ^{2} - 1} d v

因式分解

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

= 7 \cdot 8 \cdot \frac{1}{28} \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

= 7 \cdot 8 \cdot \frac{1}{28} (- \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}

= 7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x - \frac{5}{4} ) - 1}{2}))

化简

7 \cdot 8 \cdot \frac{1}{28} (- (\frac{ln \frac{4}{7} ( x - \frac{5}{4} ) + 1}{2} - \frac{ln \frac{4}{7} ( x - \frac{5}{4} ) - 1}{2})) : - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x

= - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x

解答补常数

= - ln \frac{2}{7} + \frac{4}{7} x + ln - \frac{12}{7} + \frac{4}{7} x + C

作图

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