integral of 1/(sqrt(4x^2-16))

解答

\int \frac{1}{4 x ^{2} - 16} d x

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

求解步骤

\int \frac{1}{4 x ^{2} - 16} d x

使用三角换元法

= \int \frac{sec ( u )}{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 sec (u) d u

使用积分常见定则:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

u = arcsec (\frac{2}{4} x)

代回

= \frac{1}{2} ln tan (arcsec (\frac{2}{4} x)) + sec (arcsec (\frac{2}{4} x))

化简

\frac{1}{2} ln tan (arcsec (\frac{2}{4} x)) + sec (arcsec (\frac{2}{4} x)) : \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})

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

解答补常数

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

\int \frac{2}{x ^{2} - 4 x + 13} d x

n = 0 \sum \infty \frac{1}{n ^{4}}

\frac{d y}{d x} + 3 x^{2} y = 18 x^{2}, y (0) = 8

d er i v a t i v e 38 (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x)

n = 1 \sum \infty 0. 3^{n}