Question

请问这个定积分怎么求?能写出详细过程吗谢谢～

Answer 1

let
x=√2tanu
dx=√2(secu)^2 du
x=0, u=0
x=1, u=arctan(√2/2)
∫(0->1) √(x^2+2) dx
=∫(0->arctan(√2/2)) √2(secu) . [√2(secu)^2 du]
=2 ∫(0->arctan(√2/2)) (secu)^3 du
=2(3/4 +(3/4)ln2 )
=(3/2)( 1+ ln2)
-------
∫(0->arctan(√2/2)) (secu)^3 du
=∫(0->arctan(√2/2)) (secu) dtanu
=[secu.tanu]|(0->arctan(√2/2)) - ∫(0->arctan(√2/2)) secu. (tanu)^2 du
=2/3 - ∫(0->arctan(√2/2)) secu. [(secu)^2 -1 ] du

2∫(0->arctan(√2/2)) (secu)^3 du =2/3 + ∫(0->arctan(√2/2)) secu du
∫(0->arctan(√2/2)) (secu)^3 du
=3/4 +(1/2)[ln|secu + tanu| ]|(0->arctan(√2/2))
=3/4 +(1/2)ln(2√2)
=3/4 +(3/4)ln2