SOLUTION: Please help me solve this question. integrate the following indefinite integrals: 4x-3 ---------- dx (x+1)^2

You must use partial fractions:

  4x-3         A        B
 ------- =  ------- + ----- 
 (x+1)²      (x+1)²    x+1

    4x-3 = A + B(x+1)

This must be an identity for all x

   Let x = -1

 4(-1)-3 = A + B(-1+1)

    -4-3 = A + 0

       A = -7

    4x-3 = A + B(x+1) becomes

    4x-3 = -7 + B(x+1)

   Let x = 0

4(0) - 3 = -7 + B(0+1)

      -3 = -7 + B

       4 = B

So now

  4x-3         A        B
 ------- =  ------- + ----- 
 (x+1)²      (x+1)²    x+1

becomes

  4x-3         -7       4
 ------- =  ------- + ----- 
 (x+1)²      (x+1)²    x+1


So now the problem becomes the sum of two integrals


=+ = +



Now you can finish it, right? If not post again asking how.
The fist integral is simply un and the second is 
a natural log.

Edwin