求定积分∫(x+2)dx/根号2x+1.上限4,下限0.

题目

求定积分∫(x+2)dx/根号2x+1.上限4,下限0.
需过程.谢谢!

答案

∫[(x+2)/根号2x+1]dx
= ∫{[1/2(2x+1)+2/3]/根号2x+1}dx
=∫1/2(根号2x+1)dx+∫[3/2 /根号2x+1 ]dx
=1/6 *(2x+1)^(3/2)+3/2*(根号2x+1)
上限4,下限0.
得到 =22/3