integral of (5x-4)/(sqrt(8x-5^2+9))

解答

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

\frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16 + C

求解步骤

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

乘开

\frac{5 x - 4}{8 x - 5 ^{2} + 9} : \frac{5 x}{8 x - 16} - \frac{4}{8 x - 16}

= \int \frac{5 x}{8 x - 16} - \frac{4}{8 x - 16} d x

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

= \int \frac{5 x}{8 x - 16} d x - \int \frac{4}{8 x - 16} d x

\int \frac{5 x}{8 x - 16} d x = \frac{5}{12} (x + 4) 8 x - 16

\int \frac{4}{8 x - 16} d x = 8 x - 16

= \frac{5}{12} (x + 4) 8 x - 16 - 8 x - 16

化简

\frac{5}{12} (x + 4) 8 x - 16 - 8 x - 16 : \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16

= \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16

解答补常数

= \frac{5}{12} x 8 x - 16 + \frac{2}{3} 8 x - 16 + C

\frac{d y}{d x} + y^{9} x + 4 y = 0

\int_{0}^{2} 4 x + 1 d x

x \to 0 lim (\frac{ln ( 6 x + 1 )}{x})

\int \frac{r}{4 - r ^{2}} d r

\int (5 x - 1)^{3} d x