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受欢迎的微积分 >

integral of ((8x+1))/(sqrt(4x-3))

解答

\int \frac{( 8 x + 1 )}{4 x - 3} d x

解答

\frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3 + C

求解步骤

\int \frac{8 x + 1}{4 x - 3} d x

使用换元积分法

= \int \frac{2 u ^{2} + 7}{2} 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 2 u^{2} + 7 d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 2 u^{2} d u + \int 7 d u)

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int 7 d u = 7 u

= \frac{1}{2} (\frac{2 u ^{3}}{3} + 7 u)

u = 4 x - 3

代回

= \frac{1}{2} (\frac{2 ( 4 x - 3 ) ^{3}}{3} + 7 4 x - 3)

化简

\frac{1}{2} (\frac{2 ( 4 x - 3 ) ^{3}}{3} + 7 4 x - 3) : \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3

= \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3

解答补常数

= \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3 + C

作图

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