Foo - Is this integral correct?

I'm trying to solve a differential equation but am stuck on a specific step that involves integration. Could someone please explain how the following is true?

http://img121.imageshack.us/img121/3763/integraln.jpg

If we integrate both sides both sides with respect to t, I would have expected the following result:

1/(1+t^2) - 1/(1+1^2)*y = 3*arctan(t) + C

And factoring out the 1/(1+t^2) from the left, we would get [1/(1+t^2)](y-1). Solving for y becomes

-3*arctan(t)*(1+t^2) +1 + C = y

Why am I incorrect?

I'm not 100% sure your wxy Or, xyz statement is true, but I'm not much of a math guy as far as engineers are concerned.

This part of your equation makes no sense: 1/(1+1^2). One would simply write that as 1/2.

That second 1 is a lower case T.

I'm pretty sure the answer is 42



or possibly poo

That second 1 is a lower case T.

Looks like a one to me but yeah must have been a typo.

oops, good catch. The one should have been a lower case t. :).

RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.

RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.
How can you get more general than the answer to life, the universe, and everything?

oops, good catch. The one should have been a lower case t. :).

RUOkie, 42 couldn't possibly be a valid solution since the solution is a general one -- it doesn't pertain to a specific number.

then it has to be "poo" by process of elimination. next question please.

Are you asking how the last line in the image is correct? The integral of d/dt (X) with respect to t = X (integral of a derivative of X = X). That is y/1+t^2 in your example. You already agree with the right side integral solution so now you just need to multiply both sides by 1+t^2 to solve for y.