Answer
$-2 sech \sqrt t +C$
Work Step by Step
As we are given that $\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}$
Re-write:$\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}=2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}$
Now, consider $\sqrt t =u$ and $du= \dfrac{ dt}{ 2\sqrt t}$
This implies, $2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}=2 \int (sech u \tanh u) du$
or, $=-2 (sech u) +C$
or, $=-2 sech \sqrt t +C$
