Question

求不定积分∫根号(e^x-1) dx 需要过程 谢谢

Answer 1

解：

设根号(e^x-1) =t

t^2 +1=e^x

x=ln(t^2 +1)

代入得

∫t dln(t^2 +1)

=∫2t^2/(t^2 +1) dt

=2*∫t^2/(t^2 +1) dt

=2*∫(t^2 +1-1)/(t^2 +1) dt

=2*∫[1 -1/(t^2 +1)] dt

=2*[∫1 dt -∫1/(t^2 +1) dt

=2*(t -arctant) +C(常数)

=2*【(e^x-1) -arctan(e^x-1)】+C

=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

Answer 2

解：
设根号(e^x-1) =t
t^2 +1=e^x
x=ln(t^2 +1)
代入得
∫t dln(t^2 +1)
=∫2t^2/(t^2 +1) dt
=2*∫t^2/(t^2 +1) dt
=2*∫(t^2 +1-1)/(t^2 +1) dt
=2*∫[1 -1/(t^2 +1)] dt
=2*[∫1 dt -∫1/(t^2 +1) dt
=2*(t -arctant) +C(常数)
=2*【(e^x-1) -arctan(e^x-1)】+C
=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)