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受欢迎的微积分 >

integral of 1/((2x-1)sqrt((2x-1)^2-4))

解答

\int \frac{1}{( 2 x - 1 ) ( 2 x - 1 ) ^{2} - 4} d x

解答

\frac{1}{4} arcsec (x - \frac{1}{2}) + C

求解步骤

\int \frac{1}{( 2 x - 1 ) ( 2 x - 1 ) ^{2} - 4} d x

使用换元积分法

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

提出常数:

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

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

使用三角换元法

= \frac{1}{2} \cdot \int \frac{1}{2} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 d v

常数积分:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

代回

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x - 1))

化简

\frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x - 1)) : \frac{1}{4} arcsec (x - \frac{1}{2})

= \frac{1}{4} arcsec (x - \frac{1}{2})

解答补常数

= \frac{1}{4} arcsec (x - \frac{1}{2}) + C

作图

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