Question

What is the integral of 1/(1+4y^2)?
I am just not sure what to do.

Answer 1

sub, u = 2y

Answer 2

It something that turns out to be an inverse trig function, have you ever seen ivnerse trig functions with integrals before?

Answer 3

Substitute \(y=\dfrac{1}{2}\tan u\), so that \(dy=\dfrac{1}{2}\sec^2u~du\):
\[\int\frac{1}{1+4y^2}~dx~~\Rightarrow~~\int\frac{\sec^2u}{1+4\left(\frac{1}{2}\tan u\right)^2}~du=\int du=\cdots\]