integral from 0 to 1 of (1+12t)/(1+3t)

解答

\int_{0}^{1} \frac{1 + 12 t}{1 + 3 t} d t

4 - 2 ln (2)

十进制

2.61370 \dots

求解步骤

\int_{0}^{1} \frac{1 + 12 t}{1 + 3 t} d t

长除

\frac{1 + 12 t}{1 + 3 t} : 4 - \frac{3}{3 t + 1}

= \int_{0}^{1} 4 - \frac{3}{3 t + 1} d t

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_{0}^{1} 4 d t - \int_{0}^{1} \frac{3}{3 t + 1} d t

\int_{0}^{1} 4 d t = 4

\int_{0}^{1} \frac{3}{3 t + 1} d t = 2 ln (2)

= 4 - 2 ln (2)

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