Integrate

given is 2\int_{0}^{1}{(tan^{-1}x)^{2}}dx
\Rightarrow 2\int_{0}^{1}{(tan^{-1}x)(tan^{-1}x)}dx
taking first funtion any one and integrating byparts we get
\Rightarrow\left[ 2\left[tan^{-1}x\int{tan^{-1}x}dx \right]^{1}_{0}-2\int_{0}^{1}{\frac{\int tan^{-1}x}{1+x^{2}}}dx\right]

\Rightarrow 2\left[x(tan^{-1}x)^{2}-\frac{1}{2}ln(1+x^{2}) . tan^{-1}x \right]^{1}_{0} - 2\left[\int_{0}^{1}{\frac{xtan^{-1}x}{1+x^{2}}dx} -\frac{1}{2}\int_{0}^{1}{\frac{ln(1+x^{2})}{1+x^{2}}dx}\right]

I thnk now it can be done integrate \int_{0}^{1}{\frac{xtan^{-1}x}{1+x^{2}}} by using by parts and \int_{0}^{1}{\frac{ln(1+x^{2})}{1+x^{2}}dx} can be integrated

hence i guess the question is demolished isn't it

Are u sure u applied By parts correctly?jus check.

yup i got my mistake and edited that [ made a mistake in 2nd step earlier ] sry [3]

but since i typed that once
once again try karne ka man nahin hai............
If my approach is correct then please tell me..............or shuld i try this in other way

It isnt.

check wat u did after taking ln(1+x²) = t

sry again for a silly mistake .............but edited now.......
i thnk for that log part again we got to use by parts .

I just wonder if this requires expansions - expansion of ln(1+x), where x=sin2θ.

Well i havent got this one....so cant say much.

but u all know what.....my cousin asked me this one.. day before yesterday and he says this has come in 12th board exam..

but i got it as \frac{\pi }{4}\left(\pi +1 \right)-\frac{1}{2}ln2\left[\frac{\pi }{2}+1 \right]-2

@kaymant sir wuld u please elaborate a bit abt CATALAN CONSTANT. heard of it first time [7]

yes again i did
pata nahin mujhe ho kya gaya hai

Somewhere I read that this catalan const. is -\int_{0}^{1}{\frac{lnx}{1+x^2}dx}.