Q:

Quick inverse trigonometric integration question.

I am a little confused here, how does removing $1/2$ from the function to the outside of the integral get rid of the $t$ in the numerator in this problem?
$$\eqalign{
\int\dfrac t{t^4+25}dt & = \dfrac12\int\dfrac{1}{(t^2)^2+5^2}(2)\,dt \\
&= \dfrac1{10}\arctan\left(\dfrac{t^2}5\right)+C
}$$

A:

Substitute $u=t^2$, so that $du = 2t \,dt$.  In yours, the $t$ is missing, should be $(2t)$.

