Question

to malevolence19: integral (dx)/[x(x^2 + 1)^2]

Answer 1

Give me a sec to type it up :P

Answer 2

You need to write out partial fractions: the integrand equals something/x + something/(x^2+1) + something/(x^2+1)^2

Let's see what mal19 comes up with ...

Answer 3

i thought it would be partial fractions but the solutions manual has something completely different for the work, using x = tan(theta) dx = sec^2(theta)d(theta)

Answer 4

Haha, well you can do it either way honestly. I would do it with partial fractions if it was me, but thats just me i suppose

Answer 5

i'll try partial fractions, just the solutions manual threw me off when it used trig substitution

Answer 6

oh, actually, the trig substitution here is easier

Answer 7

You eventually have to use both. But I did partial fractions and then did trig sub. Tell me what you get and if you can't get it I'll help you.
For partial fractions use this:
\[\int\limits (original)=\int\limits \left[ \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2} \right]dx\]

Answer 8

dx/(x(x^2+1)^2) = du/(tan u.sec^2 u), where x = tan u
and that equals = cos^3 u/sin u

Answer 9

and that's straight forward, as cos^3 = cos - cos.sin^2

Answer 10

Okay, after the trig sub (ignoring partial fractions) you should have (as james said)\[\int\limits \frac{\cos^3(\phi) d \phi}{\sin(\phi)}=\int\limits \frac{\cos(\phi)\cos^2(\phi)d \phi}{\sin(\phi)}=\int\limits \frac{\cos(\phi)(1-\sin^2(\phi))d \phi}{\sin(\phi)}\]
Let u=sin(phi), du=cos(phi)dphi
Then you have:
\[\int\limits \frac{(1-u^2)du}{u}=\int\limits \frac{du}{u}-\int\limits u du=\ln|u|-\frac{1}{2}u^2+C\]
But u=sin(phi)
\[\ln|\sin(\phi)|-\frac{1}{2}\sin^2(\phi)+C\]
Also: \[\sin(\phi)=\frac{x}{\sqrt{x^2+1}}\]
So:
\[\ln|\frac{x}{\sqrt{x^2+1}}|-\frac{1}{2} \left( \frac{x^2}{x^2+1} \right)+C\]