integral from 1 to t of 3/(x^3)

解答

\int_{1}^{t} \frac{3}{x ^{3}} d x

3 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

求解步骤

\int_{1}^{t} \frac{3}{x ^{3}} d x

提出常数:

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

= 3 \cdot \int_{1}^{t} \frac{1}{x ^{3}} d x

使用幂法则

= 3 [- \frac{1}{2 x ^{2}}]_{1}^{t}

带入上下限计算后得:

- \frac{1}{2 t ^{2}} + \frac{1}{2}

= 3 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\int \frac{1}{9 x ^{2} + 4} d x

\int_{- x}^{x} t^{3} d t

\frac{d}{d x} (2 x^{3} - 4 x + 5)

d er i v a t i v e \frac{t}{t - 1}

(1 + cos (t)) y^{^{'}} = (1 + e - y) sin (t)