integral from 0 to x^2 of 1/(3+x^3)

解答

\int_{0}^{x^{2}} \frac{1}{3 + x ^{3}} d y

\frac{x ^{2}}{x ^{3} + 3}

求解步骤

\int_{0}^{x^{2}} \frac{1}{3 + x ^{3}} d y

常数积分:

\int a d x = a x

= [\frac{1}{3 + x ^{3}} y]_{0}^{x^{2}}

乘

\frac{1}{3 + x ^{3}} y : \frac{y}{3 + x ^{3}}

= [\frac{y}{3 + x ^{3}}]_{0}^{x^{2}}

带入上下限计算后得:

\frac{x ^{2}}{x ^{3} + 3}

= \frac{x ^{2}}{x ^{3} + 3}