Question

求∫{x*[(x^2+1)^(1/2)]}

用代换法求

∫{x*[(x^2+1)^(1/2)]}

谢谢

Answer 1

您的题目出了一点问题, 积分的变量没有,我想您漏打了dx了.

dx很重要,说明您是对x作积分,如果您不明确积分的变量的话,那么下面就无法用代换法了,

代换法就是说,把您题目中的x dx配成:

x dx = 1/2 d(x^2)

(这是因为 (x^2)` = d(x^2) / dx = 2x )

∫[(x^2+1)^(1/2)] x dx

= ∫[(x^2+1)^(1/2)] 1/2 d(x^2)
= 1/2 ∫[(x^2+1)^(1/2)] d(x^2)
再令上式中y=x^2

=1/2 ∫[(y+1)^(1/2)] dy

又因为

dy = d(y+1)

所以
上式 = 1/2 ∫[(y+1)^(1/2)] d(y+1)

再令z=y+1

上式 = 1/2 ∫Z^(1/2) dZ

= (1/2)* (z^(3/2) / (3/2))
= 1/3 * z^(3/2)
= 1/3 * (y+1)^(3/2)
= 1/3 * (x^2+1)^(3/2)

Answer 2

原式化为
(1/2)*∫(x^2+1)^(1/2)d(x^2+1)
令u=x^2+1
则原式变为(1/2)*∫u^(1/2)du=(1/3)*u^(3/2)
带回得 原式=(1/3)*(x^2+1)^(3/2)