Question

急求！高数求积分，需要步骤，肯定采纳

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

Answer 1

(17)
let
u=√x
2udu =dx
∫ arctan√x dx
=∫ arctanu du^2

= u^2. arctanu - ∫ [u^2/(1+u^2)] du
= u^2. arctanu - ∫ [1 -1/(1+u^2)] du
= u^2. arctanu - u +arctanu +C
= xarctan√x - √x +arctan√x +C
(16)
∫xe^(2x) dx
=(1/2)∫xde^(2x)
=(1/2)x.e^(2x) -(1/2)∫e^(2x) dx
=(1/2)x.e^(2x) -(1/4)e^(2x) +C
(15)
let
x=sinu
dx=cosu du
∫√(1-x^2)/x dx
=∫ [(cosu)^2/sinu ] du
=∫ [1-(sinu)^2]/sinu du
=∫ (cscu - sinu ) du
=ln|1/x - √(1-x^2)/x | + √(1-x^2) + C
(14)

∫ (tanx)^3 . (secx)^3 dx
=∫ (tanx)^2 . (secx)^2 d(secx)
=∫ [(secx)^2 -1] . (secx)^2 d(secx)
= (1/5)(secx)^5 - (1/3)(secx)^3 + C

Answer 2

如图

追答

Answer 3

如图

追答

亲(●✿∀✿●)望采纳哟～