integral from 1 to 3 of ((x^3-1)/(x^4))

解答

\int_{1}^{3} (\frac{x ^{3} - 1}{x ^{4}}) d x

ln (3) - \frac{26}{81}

十进制

0.77762 \dots

求解步骤

\int_{1}^{3} \frac{x ^{3} - 1}{x ^{4}} d x

乘开

\frac{x ^{3} - 1}{x ^{4}} : \frac{1}{x} - \frac{1}{x ^{4}}

= \int_{1}^{3} \frac{1}{x} - \frac{1}{x ^{4}} d x

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_{1}^{3} \frac{1}{x} d x - \int_{1}^{3} \frac{1}{x ^{4}} d x

\int_{1}^{3} \frac{1}{x} d x = ln (3)

\int_{1}^{3} \frac{1}{x ^{4}} d x = \frac{26}{81}

= ln (3) - \frac{26}{81}

\int_{3}^{5} 4 x^{2} d x

\int_{1}^{2} (x^{- 3}) d x

\int_{2}^{4} (16 - x^{2}) d x

\int_{0}^{1} e^{3 x^{3}} d x

\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{sin ( x )} d x