Posted by **Ash** on Thursday, November 30, 2006 at 9:26pm.

2x^3 + 3x^2 + x + 1 dx

---------------------

-2x + 1

[2x^3 + 3x^2 + x + 1]/(-2x + 1) =

ax^2 + bx + c + d/(-2x + 1) ---->

2x^3 + 3x^2 + x + 1 = (ax^2 + bx + c)*

(-2x + 1) + d

Find a, b , c and d by equating the coeffients of equal powers of x on both sides. Then you can easily calculate the integral.

Faster way:

Expand the function about its singular point at x = -1/2. Instead of a regular Taylor expansion you'll have an expansion of the form

a(x-1/2)^(-1) + b + c(x-1/2) +

d(x-1/2)

Think about how you would generalize the Taylor expansion formula to find the coeficients in this case. Note that the expansion has to stop at the quadratic term because at large distances the function grows as x^2.

Sorry, the singular is at x = 1/2, of course.

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