The Student Room Group
Reply 1
by parts...
it's
∫1 . tan-1 x dx
Reply 2
Integration by parts:

u = tan-1 x ; du/dx = 1 +
dv/dx = 1 ; v = x

∫u(dv/dx) dx = uv - ∫v(du/dx) dx

∫tan-1 x dx = x tan-1 x - ∫ x/(1+x²) dx

2x = d/dx (1+x²) ∴ ∫ x/(1+x²) dx = ½ln |1+x²| + k

∫tan-1 x dx = x tan-1 x - ½ln (1+x²) + k
Reply 3
thanks boys
Reply 4
y = arctanx
tany = x
∫tany.dx - ∫x.dx

:rofl: :cool: