Expand Integrand: Understand Int'l Calculus Concepts

[jwxie]

Is this an expand integrand?
\int 9x ^{2} / (3 - x)^{4}

I set u = ( 3 - x)
du = -1dx

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

\int 9(3-u)^{2} (u)^{4}

the answer is
(3/x - 1) ^{-3} + c
but i can't get it...

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

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

then
negative \int (2-u) u^{1/2}

it sets
u = 2 - x
du = -dx
and x = 2-u

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

 

[tiny-tim]

Hi jwxie!

(have an integral: ∫ and try using the X² tag just above the Reply box )

"expand the integrand" simply means multiply it out, so that you get a/u⁴ + b/u³ + c/u², and then integrate that

and after you've done that, you should get a polynomial plus a constant of integration … then you can subtract a multiple of u³/u³ from the constant, and complete the cube