Question

Intergral of (5x^(2) + 20x + 6)/x(x+1)^(2)
=
integral of (A/(x+1)^(2)) + (B/(x+1)) + (c/x)

so

I have

A(x+1)x + B(x+1)^(2) + C(x+1)^(2)x + C(x+1)^(2)(x+1)/x(x+1)^(2)

The Roots being x = -1, 0
So Far

I have solved for 6 = C

How do I solve for A and B?

Answer 1

Ugh I made a mistake

Answer 2

NVm the statement is

(Ax + B(x+1)x + C(x+1)^(2))/(x+1)^(2)x

Answer 3

Ok I figured it out now :)

Answer 4

I just had the initital statement wrong

Answer 5

( 5x²+ 20x + 6) /x (x+1)²
Yep, incorrect common denom.

Answer 6

C = 6
A = 9

31 = 9 + B(2) + 6(2)^(2)
-2 = 2B
B = -1

Answer 7

Yeehh, that's what I got :)

Answer 8

Thus the intergral can be split

Integral of (6/(x+1)^(2)) + Integral of (-1/(x+1)) + integral of (9/x)

Answer 9

Just in different order A = 6, B = -1, C = 9

Answer 10

Now I can just use substitution to solve really easily

Answer 11

= 6 ln |x| - ln | x + 1| - 9 / ( x + 1) + C

Answer 12

Yup got the same

Answer 13

Thanks for the help :)

Answer 14

I haven't seen you for a while ?