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Mathematics 13 Online
OpenStudy (anonymous):

Solve the integral: \[\int\limits_{-\infty}^{+\infty}(1+x^{2})/dx\]

OpenStudy (across):

dx is in the denominator?

OpenStudy (anonymous):

yeap there is a /

OpenStudy (across):

So, you're pretty much asking us to find\[\int(1+x^2)\frac{1}{dx},\]which in words translates to "the integral of 1+x^2 with respect to the inverse of x? This has got to be the first time I encounter such statement!

OpenStudy (across):

Change "the inverse of x" with "the reciprocal of x." ^^

OpenStudy (anonymous):

I did not get what you mean with "Change "the inverse of x" with "the reciprocal of x.""

OpenStudy (jamesj):

across, you're in a playful mood today. In short, this expression as you've written it is meaningless. What you almost certainly mean is \[ \int_{-\infty}^{\infty} \frac{dx}{1+x^2} \]

OpenStudy (anonymous):

That's why i did not get what she said

OpenStudy (anonymous):

yes..try as i might..this expression korcan doesnt make sense

OpenStudy (across):

What James said. Math has got to make sense. n_O

OpenStudy (anonymous):

even when it doesn't

OpenStudy (anonymous):

it does not, if it was making any sense, then I would not date with girls

OpenStudy (anonymous):

whats that supposed to mean now?

OpenStudy (jamesj):

that's one too many negations for me. Confirm what I've written is correct and then there are quite a few people here who can help you.

OpenStudy (anonymous):

We are jus moving on a bad way, from math to philasophy :)

OpenStudy (anonymous):

if james is right then the answer is pi

OpenStudy (anonymous):

Don't give me the fish, teach me how to catch some fish, so I can feed myself, so explain the solution and how can I solve this type of integration?

OpenStudy (anonymous):

if what he wrote is correct , can we use change of variable?

OpenStudy (across):

You can solve the one James wrote by looking at a table of common integrals.

OpenStudy (anonymous):

then the answer is pi

OpenStudy (anonymous):

\[\int(1+1/u^2)du\] ?

OpenStudy (anonymous):

bcz the indefinit integral if dx/(1+x^2) is arctanx

OpenStudy (anonymous):

yeap tan^-1(x)

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