integral from-1 to 1 of 2/(sqrt(x+1))

解答

\int_{- 1}^{1} \frac{2}{x + 1} d x

42

十进制

5.65685 \dots

求解步骤

\int_{- 1}^{1} \frac{2}{x + 1} d x

提出常数:

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

= 2 \cdot \int_{- 1}^{1} \frac{1}{x + 1} d x

使用换元积分法

= 2 \cdot \int_{0}^{2} 2 d u

常数积分:

\int a d x = a x

= 2 [2 u]_{0}^{2}

带入上下限计算后得:

22

= 2 \cdot 22

化简

= 42

a re a x^{2} - 3, y = 0, - 2, 2

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{7}{2}}} d x

d er i v a t i v e f (x) = 5 sin (x) cos (x)

(x, y) \to (π, 3) lim (\frac{y cos ( 2 x )}{y + cos ( x )})

\frac{d}{d x} (m x + d)