integral from 0 to 5 of 1/(sqrt(9+4x^2))

解答

\int_{0}^{5} \frac{1}{9 + 4 x ^{2}} d x

\frac{1}{2} ln (\frac{10 + 109}{3})

十进制

0.95944 \dots

求解步骤

\int_{0}^{5} \frac{1}{9 + 4 x ^{2}} d x

使用三角换元法

= \int_{0}^{a r c t a n (\frac{10}{3})} \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_{0}^{a r c t a n (\frac{10}{3})} 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) ∣]_{0}^{a r c t a n (\frac{10}{3})}

带入上下限计算后得:

ln (\frac{10}{3} + \frac{109}{3})

= \frac{1}{2} ln (\frac{10}{3} + \frac{109}{3})

化简

= \frac{1}{2} ln (\frac{10 + 109}{3})