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受欢迎的微积分 >

integral of 4/(5x^2-3x+10)

解答

\int \frac{4}{5 x ^{2} - 3 x + 10} d x

解答

\frac{8}{191} arctan (\frac{10 x - 3}{191}) + C

求解步骤

\int \frac{4}{5 x ^{2} - 3 x + 10} d x

提出常数:

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

= 4 \cdot \int \frac{1}{5 x ^{2} - 3 x + 10} d x

对

5 x^{2} - 3 x + 10

配方:

5 (x - \frac{3}{10})^{2} + \frac{191}{20}

= 4 \cdot \int \frac{1}{5 ( x - \frac{3}{10} ) ^{2} + \frac{191}{20}} d x

使用换元积分法

= 4 \cdot \int \frac{20}{100 u ^{2} + 191} d u

提出常数:

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

= 4 \cdot 20 \cdot \int \frac{1}{100 u ^{2} + 191} d u

使用换元积分法

= 4 \cdot 20 \cdot \int \frac{1}{10 191 ( 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 20 \cdot \frac{1}{10 191} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot 20 \cdot \frac{1}{10 191} arctan (v)

代回

= 4 \cdot 20 \cdot \frac{1}{10 191} arctan (\frac{10}{191} (x - \frac{3}{10}))

化简

4 \cdot 20 \cdot \frac{1}{10 191} arctan (\frac{10}{191} (x - \frac{3}{10})) : \frac{8}{191} arctan (\frac{10 x - 3}{191})

= \frac{8}{191} arctan (\frac{10 x - 3}{191})

解答补常数

= \frac{8}{191} arctan (\frac{10 x - 3}{191}) + C

作图

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