Answer
\[ = - 2\,\,\left[ {1 + 4\ln \,\left( {\frac{3}{4}} \right)} \right]\]
Work Step by Step
\[\begin{gathered}
\int_0^1 {\frac{{dp}}{{4 - \sqrt p }}} \hfill \\
\hfill \\
substituting\,\,p = {t^2}\, \to \,\,dp = 2tdt \hfill \\
\hfill \\
= \int_0^1 {\frac{{2tdt}}{{4 - t}}} \hfill \\
\hfill \\
pull\,\,out\,\,the\,\,{\text{constant}} \hfill \\
\hfill \\
= 2\int_0^1 {\frac{{tdt}}{{4 - t}}} \hfill \\
\hfill \\
{\text{using}}\,\,{\text{the long division}} \hfill \\
\hfill \\
= - 2\int_0^1 {\,\left( {1 + \frac{4}{{t - 4}}} \right)dt} \hfill \\
\hfill \\
\,integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= - 2\,\,\left[ {t + 4\ln \left| {t - 4} \right|} \right]_0^1 \hfill \\
\hfill \\
= - 2\,\,\left[ {1 + 4\ln 3 - 0 - 4\ln 4} \right] \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
= - 2\,\,\left[ {1 + 4\ln \,\left( {\frac{3}{4}} \right)} \right] \hfill \\
\end{gathered} \]
