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受欢迎的微积分 >

integral of x/(x^4+2x^2+5)

解答

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

解答

\frac{1}{2 7} arctan (\frac{x ^{2}}{7}) + C

求解步骤

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

使用换元积分法

= \int \frac{1}{4 u ( 2 u + u + 5 )} 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 ( 2 u + u + 5 )} d u

使用指数法则:

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

\frac{1}{u ( 2 u + u + 5 )} = u^{- \frac{1}{2}} \frac{1}{( 2 + u + 5 )}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{2 + u + 5} d u

化简

u^{- \frac{1}{2}} \frac{1}{2 + u + 5} : \frac{1}{u ( u + 7 )}

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 7 )} d u

使用换元积分法

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 7} 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} + 7} d v

使用换元积分法

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{7 ( 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}{7} \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}{7} arctan (w)

代回

= \frac{1}{4} \cdot 2 \cdot \frac{1}{7} arctan (\frac{x ^{4}}{7})

化简

\frac{1}{4} \cdot 2 \cdot \frac{1}{7} arctan (\frac{x ^{4}}{7}) : \frac{1}{2 7} arctan (\frac{x ^{2}}{7})

= \frac{1}{2 7} arctan (\frac{x ^{2}}{7})

解答补常数

= \frac{1}{2 7} arctan (\frac{x ^{2}}{7}) + C

作图

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