Question

求大神帮我采纳

Answer 1

(22)
5-2x-x^2 = 6-(x+1)^2
let
x+1 =√6 sinu
dx=√6 cosu du
∫dx/√(5-2x-x^2)
=∫ du
=u + C
=arcsin[(x+1)/√6] +C
(24)
x^2+2x+10
=(x+1)^2 +9
let
x+1 = 3tanu
dx = 3(secu)^2 du
∫(2x+5)/(x^2+2x+10) dx
= ∫(2x+2)/(x^2+2x+10) dx + 3∫dx/(x^2+2x+10)
= ln|x^2+2x+10| + 3∫dx/(x^2+2x+10)
= ln|x^2+2x+10| + 3∫ 3(secu)^2 du/[9 (secu)^2 ]
= ln|x^2+2x+10| + ∫ du
= ln|x^2+2x+10| + u + C
= ln|x^2+2x+10| + arctan[(x+1)/3] + C

(26)
1/[x(2x-3)] ≡ A/x+ B/(2x-3)
=>
1 ≡ A(2x-3)+ Bx
x=0, => A = -1/3
x=3/2, => B = 2/3
∫dx/[x(2x-3)]
=∫ { -(1/3)(1/x) + (2/3) [ 1/(2x-3)] } dx
=-(1/3)lnx +(1/3)ln|2x-3| + C

追问

答案对的，但还是手写的好点，容易看懂