# Expand integrand

1. Dec 8, 2009

### 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...

Last edited: Dec 8, 2009
2. Dec 8, 2009

### 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

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