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受欢迎的微积分 >

integral of x/(x^4-13x^2+36)

解答

\int \frac{x}{x ^{4} - 13 x ^{2} + 36} d x

解答

\frac{1}{2 23} arctan (\frac{x ^{2}}{23}) + C

求解步骤

\int \frac{x}{x ^{4} - 13 x ^{2} + 36} d x

使用换元积分法

= \int \frac{1}{4 u ( u + 36 - 13 u )} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 36 - 13 u )} d u

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ( u + 36 - 13 u )} = u^{- \frac{1}{2}} \frac{1}{( u + 36 - 13 )}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 36 - 13} d u

化简

u^{- \frac{1}{2}} \frac{1}{u + 36 - 13} : \frac{1}{u ( u + 23 )}

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 23 )} d u

使用换元积分法

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 23} d v

提出常数:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 23} d v

使用换元积分法

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{23 ( w ^{2} + 1 )} d w

提出常数:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{23} \cdot \int \frac{1}{w ^{2} + 1} d w

使用积分常见定则:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (w)

代回

= \frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (\frac{x ^{4}}{23})

化简

\frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (\frac{x ^{4}}{23}) : \frac{1}{2 23} arctan (\frac{x ^{2}}{23})

= \frac{1}{2 23} arctan (\frac{x ^{2}}{23})

解答补常数

= \frac{1}{2 23} arctan (\frac{x ^{2}}{23}) + C

作图

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