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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?

$\eqalign{ \int\dfrac t{t^4+25}dt & = \dfrac12\int\dfrac{1}{(t^2)^2+5^2}(2)\,dt \\ &= \dfrac1{10}\arctan\left(\dfrac{t^2}5\right)+C }$

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    It doesn't. The second expression should have a $(2t)\;dt$ not $(2)\;dt$.2011-06-28

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

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    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?2011-06-28
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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 $\frac{1}{5} \operatorname{arctan}(\frac{u}{5})$: $= \frac{1}{10} \operatorname{arctan}(\frac{u}{5})+C$ Substitute back for $u = t^2$: $= \frac{1}{10} \operatorname{arctan}(\frac{t^2}{5})+C$

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    @Matt: The answer you quoted seems to agree with this one. Presumably the missing $t$ in the solution that you quote is a typo.2011-06-28