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受欢迎的微积分 >

integral of (2x)/((x^2+1)^2-x^2)

解答

\int \frac{2 x}{( x ^{2} + 1 ) ^{2} - x ^{2}} d x

解答

arctan (x^{2}) + C

求解步骤

\int \frac{2 x}{( x ^{2} + 1 ) ^{2} - x ^{2}} d x

提出常数:

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

= 2 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{2} - x ^{2}} d x

乘开

(x^{2} + 1)^{2} - x^{2} : x^{4} + x^{2} + 1

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

使用换元积分法

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

提出常数:

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

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

使用指数法则:

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

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

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

化简

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

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

使用换元积分法

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

提出常数:

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

= 2 \cdot \frac{1}{4} \cdot 2 \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)

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

代回

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

化简

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

= arctan (x^{2})

解答补常数

= arctan (x^{2}) + C

作图

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