## Thomas' Calculus 13th Edition

$-2 sech \sqrt t +C$
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, $=-2 sech \sqrt t +C$