$\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \int_{ 1 }^{ u } \ln\left({5+{t}^{2}}\right) \mathrm{d} t \right)$
Use the chain rule to find the derivative$\frac{ \mathrm{d} }{ \mathrm{d}u} \left( \int_{ 1 }^{ u } \ln\left({5+{t}^{2}}\right) \mathrm{d} t \right) \times \frac{ \mathrm{d}^u }{ \mathrm{d}^x }$
To find the derivative, apply the second fundamental theorem of calculus$\ln\left({5+{u}^{2}}\right) \times \frac{ \mathrm{d}^u }{ \mathrm{d}^x }$
Rewrite the derivative$\ln\left({5+{u}^{2}}\right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( u \right)$
Substitute back $u={x}^{2}$$\ln\left({5+{\left( {x}^{2} \right)}^{2}}\right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( {x}^{2} \right)$
Find the derivative$\ln\left({5+{\left( {x}^{2} \right)}^{2}}\right) \times 2x$
Simplify the expression$2\ln\left({5+{x}^{4}}\right) \times x$