Evaluate this Integral, please?

Question

∫xdx/x²-2x-3

Do it step by step, please

Thanks in advance

If you can't read it then please look at this

http://img119.imageshack.us/img119/4732/26427339lw3.jpg

Answer 1

∫x/x²-2x-3 . dx
=∫x/(x-3)(x+1).dx
=∫4x/4(x-3)(x+1).dx
=∫(3x+x)/4(x-3)(x+1).dx
=(1/4)∫(3x+3+x-3)/ (x-3)(x+1).dx
=(1/4)[∫3(x+1)/(x-3)(x+1) .dx+∫(x-3)/(x-3)(x+1).dx]
=(1/4)[∫3/(x-3).dx+ ∫1/(x+1).dx]
=(1/4)[3 log(x-3)+log(x+1)] + c


u may ask why I multiplied by 4 & put +/-3 , simply in general if we have x/(x+a)(x+b)-->multiply above & below by (a-b) 1-(-3)=1+3=4
then add above +/- ab +/-3*1=+/-3
this make u divide the equation into 2 parts

Answer 2

I = ∫ x / (x - 3)(x + 1) dx
x / (x - 3)(x + 1) = A / (x - 3) + B / (x + 1)
x = A(x + 1) + B(x - 3)
1 = A + B
0 = A - 3B
1 = 4B
B = 1/4

0 = A - 3B
A = 3/4
I = (3/4) ∫ 1 / (x - 3) dx + (1/4) ∫ 1 / (x + 1) dx
I = (3/4) log(x - 3) + (1/4) log (x + 1) + C

Answer 3

Use partial fractions.
1) Factor out x^2 - 2x - 3.
... read your textbook.

Answer 4

well dat is bit difficult

Answer 5

huh?

Answer 6

please rewrite the question.. cant read!