# Initial value problem using partial fractions

1. Apr 5, 2008

### gkchristopher

(t+1) dx/dt = x^2 + 1 (t > -1), x(0) = pi/4

I have attempted to work this by placing like terms on either side and then integrating.

1/(x^2 + 1) dx = 1/(t + 1) dt

arctan x = ln |t + 1| + C

x = tan (ln |t + 1|) + C

pi/4 = tan(ln |0 + 1|) + C

pi/4 = C

x = tan (ln |t + 1|) + pi/4

Is this even close??
This was supposed to be a partial fractions exercise but I'm not seeing how. Thanks for any help.

2. Apr 5, 2008

### Dick

No, I don't see any need for partial fractions. For the record, you can integrate 1/(x^2+1) by factoring x^2+1=(x+i)(x-i) and get an expression involving complex logs that is equivalent to arctan. But I don't know why you would want to.