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Has anyone an idea how to start with this one?
[tex]
\int_0^1\ \frac{\arctan(x)}{x(x^2+1)}\ \mbox{d}x
[/tex]
[tex]
\int_0^1\ \frac{\arctan(x)}{x(x^2+1)}\ \mbox{d}x
[/tex]
Thanks!Gib Z said:Hello Dirk, and welcome to Physicsforums!
Okay here we go:As for that integral, a good step might be to let arctan x = u, then it makes the integral look much more approachable, you might have even seen it before.
Which part, the u or the tangens?Integration by parts is the way to go for that new one.
You're right! Are at least the boundaries correctly calculated?snipez90 said:I think the integrand should be u*cot(u), since the x is in the denominator.
That's clear.Also, if you didn't know the derivative of arctan(x), the substitution x = tan(t) (or u) works just as well provided you know trig (it's essentially the same substitution).
Never heard of Lipet before!Use LIPET (log, inverse trig, polynomial, exponential, trig function) to determine which function should be "u" when you integrate by parts.
So we get:[tex] \int_0^{\pi /4} u \cdot \cot(u)\ \mbox{d}u = u \cdot \ln( \vert \sin(u) | ) |_{0}^{ \pi /4} - \int_{0}^{\pi /4} \ln(| \sin(u) | )\ \mbox{d}u [/tex]Since we have a polynomial function, let that be "u" and let the cot function be "dv".
Gib Z said:Ahh Sorry guys, my very bad mistake :( It seems x cot x has no elementary derivative >.< And the original substitution doesn't have any obvious substitution to me, It must be something nice with the bounds. Sorry again!
snipez90 said:dirk, what is the source of this problem?
exk said:I am not sure that the integral you linked is related to the problem your friend gave. Where did your friend find it?
The method for solving this integral is called partial fraction decomposition, where we break down the fraction into simpler fractions that can be integrated separately.
This integral is considered difficult because it contains a trigonometric function (arctan) and a polynomial in the denominator, which makes it challenging to integrate using basic integration techniques.
Yes, this integral can also be solved using substitution, where we substitute x with another variable to simplify the integral.
The limits of integration determine the range over which the integral is evaluated. In this case, the limits of 0 and 1 indicate that we are finding the area under the curve of the given function between x=0 and x=1.
Yes, this integral can be used to solve problems in physics, engineering, and finance, where it can help calculate areas, volumes, or even probabilities of certain events.