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`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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