Answer
$$\int {\frac{{3{x^2}}}{{2\sqrt {{x^3} + 5} }}} dx = \sqrt {{x^3} + 5} + C$$
Work Step by Step
$$\eqalign{
& {\text{Let }}\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] \cr
& {\text{use the property }}\sqrt {f\left( x \right)} = {\left[ {f\left( x \right)} \right]^{1/2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] = \frac{d}{{dx}}\left[ {{{\left( {{x^3} + 5} \right)}^{1/2}}} \right] \cr
& {\text{By the chain rule}}{\text{,}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] = \frac{1}{2}{\left( {{x^3} + 5} \right)^{1/2 - 1}}\left( {{x^3} + 5} \right)' \cr
& {\text{Differentiating}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] = \frac{1}{2}{\left( {{x^3} + 5} \right)^{ - 1/2}}\left( {3{x^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] = \frac{{3{x^2}}}{{2{{\left( {{x^3} + 5} \right)}^{1/2}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left[ {\sqrt {{x^3} + 5} } \right] = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 5} }} \cr
& {\text{Thus}}{\text{,}} \cr
& \,\,\,\,\,\,\,\,\,\,\,d\left[ {\sqrt {{x^3} + 5} } \right] = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 5} }}dx \cr
& {\text{A corresponding integration formula is}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\int {\frac{{3{x^2}}}{{2\sqrt {{x^3} + 5} }}} dx = \sqrt {{x^3} + 5} + C \cr} $$
