Question

Integrate it .
Question is in this given attachment .

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Answer 1

logx = u

x = e^u

dx = e^u du

log(logx) = log(loge^u) = log(u)

Now

d log(u) / du = 1/u

Also

1/(logx)^2 = 1/u^2

Now

Integral [ log(logx) + 1/(logx)^2 ] dx

= [ log(u) + 1/u^2 ] e^u du

= [ log(u)e^u + e^u1/u^2 ] du

= [ log(u)e^u - e^u 1/u ]

Where u = logx

= x [ log(logx) - 1/logx ] + C

Answer 2

★彡( ₳₦₴₩ɆⱤ )彡★

Before nd after ✌️

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