Answer
$-2 sech \sqrt t +C$
Work Step by Step
Given: $\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}$
This can be re-written as: $\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}=2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}$
Plug in $\sqrt t =u$ and $du= \dfrac{ dt}{ 2\sqrt t}$
Thus, $2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}=2 \int (sech u \tanh u) du=-2 sech u +C=-2 sech \sqrt t +C$
