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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

Plugging back t=\tan^{-1}(x) gives
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 .