Take the 2-minute tour ×  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?  enter image
description here  share|improve this question It doesn't. The
second expression should have a $(2t)\;dt$ not $(2)\;dt$. – 
wckronholm Jun 28 '11 at 16:42 add comment  2 Answers  up vote 4
down vote accepted  Substitute $u=t^2$, so that $du = 2t \,dt$.
In yours, the $t$ is missing, should be $(2t)$.  share|improve
this answer This is from my textbook. The textbook is wrong? – 
Matt Jun 28 '11 at 16:43 If it has $(2)$ and not $(2t)$, then
yes, it is wrong. Most textbooks nowadays have a web site of
errata. Why not see if this is listed? –  GEdgar Jun 28 '11 at
18:02 add comment  You do it like this: $$\int \frac{t}{t^4+25}
dt$$ For the integrand $\frac{t}{t^4+25}$, substitute $u = t^2$
and $du = 2 t dt$: $$= \frac{1}{2} \int \frac{1}{u^2+25} du$$
The integral of $\frac{1}{u^2+25}$ is $\frac{1}{5}
\operatorname{arctan}(\frac{