`int(sqrt(16+(logx)^(2)))/(x)dx=?`

1.	`int(sqrt(16+(logx)^(2)))/(x)dx=?`
Answer» Putting log `x=t and (1)/(x)dx=dt` we get `I=intsqrt(16+t^(2))dt=intsqrt(4^(2)+t^(2))dt` `=(t)/(2)sqrt(16+t^(2))+(16)/(2)log\|t+sqrt(16+t^(2))\|+C` `=(1)/(2)logx*sqrt(16+(logx)^(2))+8 log\|logx+sqrt(16+(logx)^(2))\|+C`.

Discussion

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