Very (I'm sure) simple integration question

amacachi

This basically: http://integrals.wolfram.com/index.jsp?expr=2x%2F%28x%5E2+%2B+1%29%5E2&random=false

I just can't see how to get to the answer given. I can see how differentiating the answer gives the question but I can't see how the integration is done. Anyone feeling generous?

Michael Collins

Substitution should do it:

u = x^2 + 1

amacachi

Ugh, did that then increased the power by going from -2 to -3. Not looking good for tomorrow.:o

2manyconditions

amacachi wrote: »

Ugh, did that then increased the power by going from -2 to -3. Not looking good for tomorrow.:o

Not that I know anything about integration but shouldn't you be going the opposite, ie from -2 to -1.

I haven't done integration in years so I was trying to solve this one myself but I cant figure out what happened to the 2x over the line?

amacachi

2manyconditions wrote: »

Not that I know anything about integration but shouldn't you be going the opposite, ie from -2 to -1.

I haven't done integration in years so I was trying to solve this one myself but I cant figure out what happened to the 2x over the line?

Yeah that's the mistake I meant, I "raised" the power from -2 to -3 which messed up my answer.

Using the u=x^2 +1 ya get du=2xdx so it's just du/u^2.

Eliot Rosewater

In future it'd be worth giving Wolfram Alpha the integral, because there is some scope for seeing how to do it.

Your particular question, solved. Next to the answer there's a link "Show steps" which can be handy for figuring out how to solve the problem some times.


I say some times because Wolfram Alpha is a computer program, and it integrates in the way computers do most reliably: through brute force. Thus, you may put a relatively easy integral into Alpha and find that it uses substitution, like, 10 times. So it won't help you with every problem. That's what boards.ie maths forum is for.

amacachi

Another question, I need to find a general solution to this initial value problem:

x(dy/dx) = (cosx/x^2) - 3y

I'm guessing there's something more to it than just rearranging it to get all x's on one side and all y's on the other. If that's all that has to be done I'm stumped. :pac:

Fremen

One trivial step.
An integrating factor
solves your conundrum.

...and yes, that was a haiku.

Sev

Try multiplying both sides of the equation by x^2, then see if you can simplify the expression using the product rule.

amacachi

Ah well, too late now, don't have to worry about maths again for at least 3 years!

Michael Collins

amacachi wrote: »

Ah well, too late now, don't get to worry about maths again for at least 3 years!

FYP.