Sunday

March 1, 2015

March 1, 2015

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

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

-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.

**Answer this Question**

**Related Questions**

Math...For All Math Tutors - I would like to know how many of the people ...

math - all i can say is math stinx dont mess with math. without math, a lot of ...

math - I just do not understand how to figure this out. Find the number of ...

English Grammar - A: What time is your math/Math class? B: At eleven forty-five...

Math - Is IB Math Studies the same as Pre-Calculus? I heard that IB Math Studies...

3rd grade Math - Actual question on homework."EXPLAIN HOW YOU CAN BREAK APART ...

math math math math - a. what is the general term of the following series? 60/...

What Makes A Good Math Teacher? - I think a good math teacher is someone who can...

Math - All of my life I have gotten a C average in math. I have been trying to ...

English - Revise the folloing sentences to create parallel constructions. (...