# Quiz question that stumped me

• Sep 12th 2008, 10:47 AM
freyrkessenin
Quiz question that stumped me
We had a quiz today and there was a problem where I didn't even know where to start... I was hoping someone could help me out on it.

I'm not very good with the syntax to this editor, but I'll try my best.

Evaluate the following integral:

∫ x^4 / (x^10 + 16) dx

Clarification:
Numerator: x^4
Denominator: x^10 + 16

(My problem was I didn't know if I should try substitution, because it doesn't look like I can factor the denominator.)

Thanks,
-Freyr.
• Sep 12th 2008, 11:18 AM
Soroban
Hello, Freyr!

Quote:

$\int\frac{x^4}{x^{10} + 16}\,dx$

We have: . $\int \frac{x^4\,dx} {(x^5)^2 + 16}$

Let $u = x^5\quad\Rightarrow\quad du = 5x^4\,dx \quad\Rightarrow\quad x^4\,dx = \frac{du}{5}$

Substitute: . $\int\frac{\frac{du}{5}}{u^2 + 16} \;=\;\frac{1}{5}\int\frac{du}{u^2+16} \quad\hdots$ Got it?

• Sep 12th 2008, 12:59 PM
freyrkessenin
Oh... Wow. That's really thinking outside the box. But when you do it, it looks so simple.

So the integral would be 1/5 [ ln | u^2 + 16| ] ?
• Sep 12th 2008, 03:22 PM
ILoveMaths07
Integrals
NEVER use $\ln$ when the denominator contains a quadratic expression. Use $\ln$ only if it's LINEAR. You can always check your answer by differentiation.

Recall that an integral of the type $\int\frac{dx}{a^2+x^2}$ can be simplified by letting $x = a\ tan\ (\theta)$. So in the end, the integral becomes $= \frac{1}{a} \arctan \frac{x}{a} + C.$

And note that, in this case, $a = 4, x = u$. Either way, you're substituting twice, so do not use the variable $x$. Make sure your final answer contains $x$ and not $u$ or any other variable.

I hope that helps. :)

P.S. Maths ALWAYS requires you to think 'out of the box' :p. That's why it's so much fun.
• Sep 12th 2008, 06:32 PM
mr fantastic
Quote:

Originally Posted by ILoveMaths07
NEVER use $\ln$ when the denominator contains a quadratic expression. Use $\ln$ only if it's LINEAR. You can always check your answer by differentiation.

[snip]

So what would you suggest the answer to $\int \frac{2x}{x^2 + 1} \, dx$ is equal to!?

I understand what you're trying to say. But such sweeping generalisations are rarely true and serve only to create confusion.
• Sep 13th 2008, 05:11 PM
ILoveMaths07
Heh, that would be $ln |x^{2}+1| + C$. Umm... Then how would you generalise what I was trying to say? How would you put it in other words? (Nerd)

ILoveMaths07.
• Sep 13th 2008, 06:53 PM
mr fantastic
Quote:

Originally Posted by ILoveMaths07
Heh, that would be $ln |x^{2}+1| + C$. Umm... Then how would you generalise what I was trying to say? How would you put it in other words? (Nerd)

ILoveMaths07.

You can't, unless you want a cookbook that rivals the American constitution for complexity.

There are general principles but ultimately the student's ability and experience is what solves the problem. Especially experience.

It also helps if the student has met and can remember the basic standard forms (like the one that appears in this thread). Unfortunately that's rarely the case.

The mistake the OP made is a consequence of misunderstanding what is in the Cook Book Calculus For Dummies.