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受欢迎的微积分 >

integral of 1/(4u^2+9-41u)

解答

\int \frac{1}{4 u ^{2} + 9 - 41 u} d u

解答

- \frac{1}{1537} (ln \frac{8 u - 41}{1537} + 1 - ln \frac{8 u - 41}{1537} - 1) + C

求解步骤

\int \frac{1}{4 u ^{2} + 9 - 41 u} d u

对

4 u^{2} + 9 - 41 u

配方:

4 (u - \frac{41}{8})^{2} - \frac{1537}{16}

= \int \frac{1}{4 ( u - \frac{41}{8} ) ^{2} - \frac{1537}{16}} d u

使用换元积分法

= \int \frac{16}{64 v ^{2} - 1537} d v

提出常数:

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

= 16 \cdot \int \frac{1}{64 v ^{2} - 1537} d v

使用换元积分法

= 16 \cdot \int \frac{1}{8 1537 ( w ^{2} - 1 )} d w

提出常数:

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

= 16 \cdot \frac{1}{8 1537} \cdot \int \frac{1}{w ^{2} - 1} d w

因式分解

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

= 16 \cdot \frac{1}{8 1537} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

提出常数:

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

= 16 \cdot \frac{1}{8 1537} (- \int \frac{1}{- w ^{2} + 1} d w)

使用积分常见定则:

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

= 16 \cdot \frac{1}{8 1537} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代回

= 16 \cdot \frac{1}{8 1537} - \frac{ln \frac{8}{1537} ( u - \frac{41}{8} ) + 1}{2} - \frac{ln \frac{8}{1537} ( u - \frac{41}{8} ) - 1}{2}

化简

16 \cdot \frac{1}{8 1537} - \frac{ln \frac{8}{1537} ( u - \frac{41}{8} ) + 1}{2} - \frac{ln \frac{8}{1537} ( u - \frac{41}{8} ) - 1}{2} : - \frac{1}{1537} (ln \frac{8 u - 41}{1537} + 1 - ln \frac{8 u - 41}{1537} - 1)

= - \frac{1}{1537} (ln \frac{8 u - 41}{1537} + 1 - ln \frac{8 u - 41}{1537} - 1)

解答补常数

= - \frac{1}{1537} (ln \frac{8 u - 41}{1537} + 1 - ln \frac{8 u - 41}{1537} - 1) + C

作图

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