Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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
1  
It doesn't. The second expression should have a $(2t)\;dt$ not $(2)\;dt$. – wckronholm Jun 28 '11 at 16:42

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
up vote 1 down vote

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 $u = t^2$: $$= \frac{1}{10} \operatorname{arctan}(\frac{t^2}{5})+C$$

share|improve this answer
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.