Question

integration [log(logx) + 1/(logx)2] DX​

Answer 1

Step-by-step explanation:

take log x= t

so x= e^t

dx=e^tdt

integration (log t + 1/t^2) et dt

et (log t + 1/t - 1/t + 1/t^2) dt. (add and subtract 1/t)

et(log t +1/t) + et(-1/t + 1/t^2)

using--integration of ex(fx'+fx) = ex(fx) +C

et(logt) + et(-1/t) + c

putting values back that we substituted

xlog(logx) - x/logx + c