Question

求一道定积分问题（高分求助）

求一函数的定积分在线等！被积函数为r^3*(1+r^2)^1/2积分上下限分别为2^1/2和0，要求写出详细步骤！手机只能加20分，答对追加50分！

Answer 1

令1+x^2=y

∫x^3•(1+x^2)^(1/2)dx
=(1/2)•∫x^2•(1+x^2)^(1/2)d(x^2)
=(1/2)•∫(1+x^2)•(1+x^2)^(1/2)-(1+x^2)^(1/2)d(x^2)
=(1/2)•∫(1+x^2)•(1+x^2)^(1/2)d(1+x^2)-(1/2)•∫(1+x^2)^(1/2)d(1+x^2)
=(1/2)•∫y•y^(1/2)dy-(1/2)•∫y^(1/2)dy
=(1/3)•∫yd[y^(3/2)]-(1/3)•y^(3/2)
=(1/3)•y^(5/2)-(1/3)•∫y^(3/2)dy-(1/3)•y^(3/2)
=(1/3)•y^(5/2)-(2/15)•y^(5/2)-(1/3)•y^(3/2)
=(1/5)•y^(5/2)-(1/3)•y^(3/数返大2)
=(1/薯竖5)•(1+x^2)^(5/2)-(1/3)•(1+x^2)^(3/2)
=(1/5)•(1+2)^(5/2)-(1/3)•(1+2)^(3/2)-[(1/5)•(1+0)^(5/2)-(1/3)•(1+0)^(3/2)
=9√3/世坦5-√3-(1/5-1/3)
=4√3/5+2/15

Answer 2

由题意，得：
=积分号下（0，根号2）r^3根号(r^2+1)dr
=1/2*积分号下（0，简铅根号2）r^2根号(r^2+1)d(r^2)
=1/2*积分号下（0，根号2）[(r^2+1)-1]根号(r^2+1)d(r^2+1)
=1/2*积分号下（0，根号2）[(r^2+1)^(3/2)-根号(r^2+1)]d(r^2+1)
=1/2*[2(r^2+1)^(5/2)/5-2(r^2+1)^(3/2)/3](0，根号2)
=(r^2+1)^(5/2)/5-(r^2+1)^(3/2)/3](0，拦指好根号2)
=[3^(5/2)-1]/5-[3^(3/逗芦2)-1]/3
=9根号3/5-根号3+2/15
=4根号3/5+2/15

Answer 3

无理函数积分 查积分表
这类题（只要橘瞎喊可积）有普遍规律：圆野
令无理式为t（或任意参数）神历

这题，令(1+r^2)^(1/2)=t