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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{4} arctan (\frac{x ^{2}}{2}) + C

求解步骤

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

使用换元积分法

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

使用指数法则:

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

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

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

化简

u^{- \frac{1}{2}} \frac{1}{2 + u + 2} : \frac{1}{u ( u + 4 )}

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

使用换元积分法

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

使用换元积分法

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

代回

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

化简

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

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

解答补常数

= \frac{1}{4} arctan (\frac{x ^{2}}{2}) + C

作图

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