integral of 1/(4x^2+9)

解答

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

\frac{1}{6} arctan (\frac{2}{3} x) + C

求解步骤

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

使用换元积分法

= \int \frac{1}{6 ( u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

使用积分常见定则:

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

= \frac{1}{6} arctan (u)

u = \frac{2}{3} x

代回

= \frac{1}{6} arctan (\frac{2}{3} x)

解答补常数

= \frac{1}{6} arctan (\frac{2}{3} x) + C

\frac{d}{d x} (\frac{1}{6 x ^{2}})

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 5 x + 6 )} d x

\int \frac{1}{3 + 2 x} d x

\int - 1 sin^{3} (x) cos^{2} (x) d x

x \to \infty lim (x)