integral of 1/((x(sqrt((4x^2-9)))))

解答

\int \frac{1}{( x ( ( 4 x ^{2} - 9 ) ) )} d x

\frac{1}{3} arcsec (\frac{2}{3} x) + C

求解步骤

\int \frac{1}{x 4 x ^{2} - 9} d x

使用三角换元法

= \int \frac{1}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int 1 d u

常数积分:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

u = arcsec (\frac{2}{3} x)

代回

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{2}{3} x)

化简

= \frac{1}{3} arcsec (\frac{2}{3} x)

解答补常数

= \frac{1}{3} arcsec (\frac{2}{3} x) + C

x \to 5 lim (2 x - 7)

\int \frac{1}{π} x cos (n x) d x

\int sin (8 x) cos (2 x) d x

d er i v a t i v e y = \frac{1}{27} (9 x^{2} + 6)^{\frac{3}{2}}

\frac{d}{d x} (tan (\frac{2}{x}))