Question

求解第11和13题？

Answer 1

(11)
let
x=2sinu
dx=2cosu du
∫dx/[x.√(4-x^2)]
=∫2cosu du/( 4 sinu.cosu )
=(1/2) ∫ cscu du
=(1/2)ln|cscu -cotu| + C
=(1/2)ln|2/x -√(4-x^2)/x | + C
(13)
2x^2+4x+5 = 2(x+1)^2 +3
let
x+1 =(√6/2) tanu
dx = (√6/2) (secu)^2 du
∫(x-2)/√(2x^2+4x+5) dx

=(1/4)∫(4x+4)/√(2x^2+4x+5) dx -3∫dx/√(2x^2+4x+5) dx
=(1/2)√(2x^2+4x+5) -3∫dx/√(2x^2+4x+5) dx
=(1/2)√(2x^2+4x+5) -3∫(√6/2) (secu)^2 du/ (√3secu)
=(1/2)√(2x^2+4x+5) -(3√2/2) ∫secu du
=(1/2)√(2x^2+4x+5) -(3√2/2) ln|secu + tanu | + C'
=(1/2)√(2x^2+4x+5) -(3√2/2) ln|√(2x^2+4x+5)/√6 + 2(x+1)/√6 | + C'
=(1/2)√(2x^2+4x+5) -(3√2/2) ln|√(2x^2+4x+5) + 2(x+1) | + C
4(x+1)^2/6 +1
=4x^2+8x+10/6
=2x^2+4x+5/6