Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{p}{{\root 5 \of {{p^2} + 1} }}dp} \cr
& {\text{definition of improper integral}} \cr
& \int_0^\infty {\frac{p}{{\root 5 \of {{p^2} + 1} }}dp} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{p}{{\root 5 \of {{p^2} + 1} }}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{{\left( {{p^2} + 1} \right)}^{ - 1/5}}pdp} \cr
& {\text{evaluate the integral}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left. {\left( {\frac{{{{\left( {{p^2} + 1} \right)}^{4/5}}}}{{4/5}}} \right)} \right|_0^b \cr
& = \frac{5}{8}\mathop {\lim }\limits_{b \to \infty } \left. {\left( {{{\left( {{p^2} + 1} \right)}^{4/5}}} \right)} \right|_0^b \cr
& = \frac{5}{8}\mathop {\lim }\limits_{b \to \infty } \left( {{{\left( {{b^2} + 1} \right)}^{4/5}} - {{\left( {{0^2} + 1} \right)}^{4/5}}} \right) \cr
& = \frac{5}{8}\mathop {\lim }\limits_{b \to \infty } \left( {{{\left( {{b^2} + 1} \right)}^{4/5}} - 1} \right) \cr
& {\text{evaluate the limit}} \cr
& = \frac{5}{8}\left( {{{\left( {{\infty ^2} + 1} \right)}^{4/5}} - 1} \right) \cr
& = \infty \cr
& {\text{diverges}} \cr} $$
