- #1

jwxie

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Is this an expand integrand?

[tex]\int 9x[/tex] [tex]^{2}[/tex] [tex]/[/tex] (3 - x)[tex]^{4}[/tex]

I set u = ( 3 - x)

du = -1dx

and so if i treat x = 3 - u , i might get this integral

[tex]\int[/tex] 9(3-u)[tex]^{2}[/tex] (u)[tex]^{4}[/tex]

the answer is

(3[tex]/[/tex]x - 1) [tex]^{-3}[/tex] + c

but i can't get it...

Originally, from the book, it gave a simple example like this

[tex]\int[/tex] [tex]x[/tex] (2-x)[tex]^{1/2}[/tex]

then

negative [tex]\int[/tex] [tex](2-u) [/tex] u[tex]^{1/2}[/tex]

it sets

u = 2 - x

du = -dx

and x = 2-u

I just don't get what EXPANDED INTEGRAND is really doing...

[tex]\int 9x[/tex] [tex]^{2}[/tex] [tex]/[/tex] (3 - x)[tex]^{4}[/tex]

I set u = ( 3 - x)

du = -1dx

and so if i treat x = 3 - u , i might get this integral

[tex]\int[/tex] 9(3-u)[tex]^{2}[/tex] (u)[tex]^{4}[/tex]

the answer is

(3[tex]/[/tex]x - 1) [tex]^{-3}[/tex] + c

but i can't get it...

Originally, from the book, it gave a simple example like this

[tex]\int[/tex] [tex]x[/tex] (2-x)[tex]^{1/2}[/tex]

then

negative [tex]\int[/tex] [tex](2-u) [/tex] u[tex]^{1/2}[/tex]

it sets

u = 2 - x

du = -dx

and x = 2-u

I just don't get what EXPANDED INTEGRAND is really doing...

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