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FinalFantasy

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want2beSMART said:
the integral of dx / [sqrt x ( 1+x) ]
let u=sqrt(1+x)
u^2=(1+x)
x=u^2-1
dx=2u du
Let I=integral of dx /sqrt x ( 1+x)
I=int. 2u du\u(sqrt(u^2-1)
=2int. du\sqrt(u^2-1)
=2 ln (u+sqrt(u^2-1) )+c
=2ln (sqrt(1+x)+sqrt(x) )+C
 

want2beSMART

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FinalFantasy said:
=2int. du\sqrt(u^2-1)
=2 ln (u+sqrt(u^2-1) )+c
=2ln (sqrt(1+x)+sqrt(x) )+C
i dont get it from here on

how come it is ln (U+sqrt(u^2-1) ) when the U was eliminated in the previous step
 

shafqat

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let u^2 = x
then 2udu = dx
so int. dx/(sqrt x)(1 + x) = int. 2udu/u(1+u^2)
= int. 2du/(1+u^2)
= 2tan-1(u)
= 2tan-1(sqrt x)
 

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