微積分-積分型態

Let x = 2tanθ, dx = 2(secθ)^2 dθ

∫ 1/[x^2√(x^2 + 4)] dx

= ∫ 2(secθ)^2/[4(tanθ)^2(2secθ)] dθ

= 1/4 ∫ (secθ)/(tanθ)^2 dθ

= 1/4 ∫ (cosθ)/(sinθ)^2 dθ

= 1/4 ∫ (sinθ)^(-2) d(sinθ)

= (-1/4)(sinθ)^(-1)

= (-1/4) √[1 + (x^2/4)]/(x/2)

= -√[1 + (x^2/4)]/(2x)

= -√(4 + x^2)/(4x)

∫ 1/[x^2*(x^2+4)^(1/2)] dx = - ∫ (1/x)/√ [1+(2/x)^2] d(1/x) = - √ [1+(2/x)^2] + C

2011-04-18 00:56:09 補充：
抱歉，係數忘了平衡

重打一次

∫ 1/[x^2*(x^2+4)^(1/2)] dx = - ∫ (1/x)/√ [1+(2/x)^2] d(1/x) = -(1/4) ∫ (2/x) / √ [1+(2/x)^2] d(2/x) = - (1/4)√ [1+(2/x)^2] + C

到底是要問幾次?
還是大家的解答都不合你意?

S 1/[x^2*(x^2+4)^(1/2)] dx =?

令x=2tan(a) ,dx=2[sec(a)]^2 da帶入原式
( tan(a)=x/2 =>
以x,2為兩股製造直角三角形,斜邊=(x^2+4)^0.5
csc(a)=(x^2+4)^0.5 / x --------------(*1) )

所求=
S 2[sec(a)]^2/{ [2 tan(a)]^2* (4*tan (a)^2+4 )^0.5 } da

(4*tan (a)^2+4 )^0.5= 2*[tan(a)^2+1]^0.5 =2*sec(a) )

所求=S 2[sec(a)]^2/{4*[tan(a)]^2*[2(sec(a)] } da

=(1/4) S sec(a) / [tan(a)]^2 da

=(1/4) S [1/cos(a)]*[cos(a)]^2/[sin(a)]^2 da

=(1/4) S [cos(a)/sin(a)]* [1/sin(a)] da

=(1/4) S cot(a)*csc(a) da

= -(1/4) csc(a) +c

=(-1/4)*[(x^2+4)^0.5 / x] +c




2011-04-17 11:40:20 補充：
d (csc (a) )/dx = -csc(a)*cot(a)

所以
-S csc(a)*cot(a) da=csc(a) +c