Partial Integration for [tex]\frac{xarctan(x)}{(1+x^2)^2}[/itex]

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How do I integrate:
\frac{xarctan(x)}{(1+x^2)^2}[/itex]
 
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You have to show us how you started first.
 
Tom Mattson said:
You have to show us how you started first.

How I started first? I tried with partial integration in many different ways.. :bugeye:
 
What do you mean by "partial integration"? I've only heard that term used in reference to integrals of functions of several variables. Did you mean to say, "integration by parts"? If so, then please show us what you did.
 
Before you integrate something, it's always good to spend some time to look at it closely.
Now if you choose u = x, and dv = (arctan(x)dx) / (1 + x2)2, then it's very hard to find v.
If you choose u = x / (1 + x2)2, and dv = arctan(x)dx, then you'll get a mess when you try to find du, and obviously, you are complicating the integrand.
And if you choose u = 1 / (1 + x2)2, and dv = x arctan(x) dx, then it's hard to find v.
...
And if you choose u = arctan(x), and dv = (x dx) / (1 + x2)2, you can make the integrand look simplier. Now just try it.
You then come up with something like:
\int \frac{dx}{(1 + x ^ 2) ^ 2}, you can again try to integrate it by parts.
Viet Dao,
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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