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受欢迎的微积分 >

integral of (-3)/(3x^2+8x+4)

解答

\int \frac{- 3}{3 x ^{2} + 8 x + 4} d x

解答

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

求解步骤

\int \frac{- 3}{3 x ^{2} + 8 x + 4} d x

提出常数:

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

= - 3 \cdot \int \frac{1}{3 x ^{2} + 8 x + 4} d x

对

3 x^{2} + 8 x + 4

配方:

3 (x + \frac{4}{3})^{2} - \frac{4}{3}

= - 3 \cdot \int \frac{1}{3 ( x + \frac{4}{3} ) ^{2} - \frac{4}{3}} d x

使用换元积分法

= - 3 \cdot \int \frac{3}{9 u ^{2} - 4} d u

提出常数:

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

= - 3 \cdot 3 \cdot \int \frac{1}{9 u ^{2} - 4} d u

使用换元积分法

= - 3 \cdot 3 \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

= - 3 \cdot 3 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} - 1} d v

因式分解

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

= - 3 \cdot 3 \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

= - 3 \cdot 3 \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}

= - 3 \cdot 3 \cdot \frac{1}{6} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

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

化简

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

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

解答补常数

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

作图

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