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2015-04-18T14:05:32+05:30

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\int\frac{x\tan^{-1}x}{\sqrt{1+x^2}}

substitute t=\tan^{-1}(x)\implies dt=\frac{dx}{1+x^2} and we get
=\int t\sec(t)\tan(t)\,dt

Next integrate by parts letting
t=u and dv=\sec(t)\tan(t),
dt=du and v=\sec(t)

=t\sec(t)-\int \sec(t)\,dt
=t\sec(t)-\ln|\sec(t)+\tan(t)|+C

Plugging back t=\tan^{-1}(x) gives
=\tan^{-1}(x)\sqrt{1+x^2}-\ln|\sqrt{1+x^2}+x|+C
2 5 2
ok got it
thats a good question, draw a right triangle based on t = tan^(-1)(x) and you will see that sect = sqrt(1+x^2)
can u solve my next q. ? i have exam tom
I'll try.. post..
posted .