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受欢迎的微积分 >

integral of 1/(x(8x+9))

解答

\int \frac{1}{x ( 8 x + 9 )} d x

解答

- \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣)) + C

求解步骤

\int \frac{1}{x ( 8 x + 9 )} d x

乘开

x (8 x + 9) : 8 x^{2} + 9 x

= \int \frac{1}{8 x ^{2} + 9 x} d x

对

8 x^{2} + 9 x

配方:

8 (x + \frac{9}{16})^{2} - \frac{81}{32}

= \int \frac{1}{8 ( x + \frac{9}{16} ) ^{2} - \frac{81}{32}} d x

使用换元积分法

= \int \frac{32}{256 u ^{2} - 81} d u

提出常数:

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

= 32 \cdot \int \frac{1}{256 u ^{2} - 81} d u

使用换元积分法

= 32 \cdot \int \frac{1}{144 ( v ^{2} - 1 )} d v

提出常数:

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

= 32 \cdot \frac{1}{144} \cdot \int \frac{1}{v ^{2} - 1} d v

因式分解

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

= 32 \cdot \frac{1}{144} \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

= 32 \cdot \frac{1}{144} (- \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}

= 32 \cdot \frac{1}{144} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( x + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( x + \frac{9}{16} ) - 1}{2}))

化简

32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( x + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( x + \frac{9}{16} ) - 1}{2})) : - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣))

= - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣))

解答补常数

= - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣)) + C

作图

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