integral from 6 to 8 of (96)/((x-6)^3)

解答

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

发散

求解步骤

\int_{6}^{8} \frac{96}{( x - 6 ) ^{3}} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 96 \cdot \int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

使用换元积分法

= 96 \cdot \int_{0}^{2} \frac{1}{u ^{3}} d u

使用幂法则

= 96 [- \frac{1}{2 u ^{2}}]_{0}^{2}

带入上下限计算后得:

发散

= 发散

\int e^{3 x} \cdot x d x

a re a y = sin^{2} (x), y = sin^{3} (x), 0 \leq x \leq π

[x^{2} + y^{2}] d x = x y d y

d er i v a t i v e - \frac{1}{( x + 1 ) ^{2}}

\int (\frac{e ^{x}}{x}) d x