# 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?

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It doesn't. The second expression should have a $(2t)\;dt$ not $(2)\;dt$. – wckronholm Jun 28 '11 at 16:42

Substitute $u=t^2$, so that $du = 2t \,dt$. In yours, the $t$ is missing, should be $(2t)$.
 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
$$\int \frac{t}{t^4+25} dt$$ 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 $\frac{1}{5} \operatorname{arctan}(\frac{u}{5})$: $$= \frac{1}{10} \operatorname{arctan}(\frac{u}{5})+C$$ Substitute back for $u = t^2$: $$= \frac{1}{10} \operatorname{arctan}(\frac{t^2}{5})+C$$
 The answer above is from the text book is their process incorrect? – Matt Jun 28 '11 at 16:58 There's obviously a typo in the text book. See wckronholm's comment – Nana Jun 28 '11 at 17:14 @Matt: The answer you quoted seems to agree with this one. Presumably the missing $t$ in the solution that you quote is a typo. – André Nicolas Jun 28 '11 at 17:15