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[(tanx)/x].dx
This can be integrated by integration by partslet 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1)
now consider the second term in the RHS∫ logx sec^2x dx, again applying by parts techniquelet u= logx , du= 1/xdx ∫sec^2x dx= dv , v= tanx on integrating
so this becomes logxtanx- ∫tanx/x dx
substituting in eqn (1)integral tanx/x dx= logxtanx+logxtan- ∫tanx/xdx2 ∫tanx/x dx= 2logxtanx
so ∫tanx/x =logx tanx+c
Ans
its a very easy question... try teaching something more useful to students....
let 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1)
substituting in eqn (1)
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