�ç 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.
To solve this problem, we need to find the integral of the given expression:
∫[(2x^3 + 3x^2 + x + 1) / (-2x + 1)] dx
First, we can try to divide the numerator by the denominator using long division. This will give us the quotient and the remainder:
2x^3 + 3x^2 + x + 1 = (-2x + 1)(ax^2 + bx + c) + d
By equating the coefficients of equal powers of x on both sides, we can find the values of a, b, c, and d.
Alternatively, we can use a faster method by expanding the function about its singular point at x = 1/2. Instead of using a regular Taylor expansion, we will have an expansion of the form:
a(x - 1/2)^(-1) + b + c(x - 1/2) + d(x - 1/2)^2
To find the coefficients in this expansion, we can generalize the Taylor expansion formula. Note that the expansion should stop at the quadratic term since the function grows as x^2 at large distances.
Apologies for the confusion in my previous response. The correct singular point is x = 1/2, not -1/2.