int 1/(t(1+(lnt)^2)) dt Use integration tables to find the indefinite integral.

Indefinite integral are written in the form of int f(x) dx = F(x) +C
 where: f(x) as the integrand
           F(x) as the anti-derivative function 
           C  as the arbitrary constant known as constant of integration
To evaluate the given integral problem: int 1/(t(1+(ln(t))^2)) dt , we may apply u-substitution by letting: u =ln(t) then du = 1/t dt .
Plug-in u =ln(t) and du= 1/tdt on int 1/(t(1+(ln(t))^2)) dt , the integral becomes:
int 1/(t(1+(ln(t))^2)) dt =int1/(1+(ln(t))^2) *1/t dt
                                   =int1/(1+u^2) du
From the integration table, we have indefinite integration formula for rational function as: int 1/(1+x^2) dx=arctan(x) +C . The int 1/(1+x^2)dx resembles the format of int 1/(1+u^2) du where  " u" corresponds to "x" .
 This is our clue that we may apply the aforementioned formula for rational function.
We get: int1/(1+u^2) du= arctan(u) +C .
 Plug-in u =ln(t) on arctan(u) +C , we get the indefinite integral as:
int 1/(t(1+(ln(t))^2)) dt= arctan(ln(t))+C

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