integral from 0 to 3 of (x^2)/(x^3+6)

解答

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 6} d x

\frac{1}{3} ln (\frac{11}{2})

十进制

0.56824 \dots

求解步骤

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 6} d x

使用换元积分法

= \int_{6}^{33} \frac{1}{3 u} 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_{6}^{33} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{6}^{33}

带入上下限计算后得:

ln (33) - ln (6)

= \frac{1}{3} (ln (33) - ln (6))

化简

= \frac{1}{3} ln (\frac{11}{2})

\int_{0}^{2 π} - 5 sin^{2} (t) d t

\int_{1}^{4} t^{2} - t - 6 d t

\int_{0}^{2} 6 x^{3} x^{4} + 9 d x

\int_{- 1}^{2} (x^{2} - 2 x + 2) d x

\int_{0}^{5} (x^{2} + 2 x - 15) d x