Calc II Tough integral problem?

Question

Well, at least i think its tough... i have NO idea where to even begin:

(e^(arctan(y)))dy / (1+y^2)

Thanks for the help

Answer 1

Actually it's very easy after you realize the little trick.

Keep in mind the derivative of arctan(v) = 1 / (1 + v^2)

So use u-substitution.

Let u = arctan(y)
Then du = dy / (1 + y^2)
Thus dy = (1 + y^2) du

So
(e^(arctan(y)))dy / (1+y^2)
= (e^(u)) du

And if you integrate that you get:
e^u + C
Which is
e^(arctan(y)) + C

Answer 2

put arc tan y = z so 1/(1+y^2)*dy =dz
so your integral becomes
Int e^z dz = e^z+C = e^arctg y+C