$\int{ \frac{ t-25 }{ 2t } } \mathrm{d} t$
Use the property of integral $\begin{array} { l }\int{ a \times f\left( x \right) } \mathrm{d} x=a \times \int{ f\left( x \right) } \mathrm{d} x,& a \in ℝ\end{array}$$\frac{ 1 }{ 2 } \times \int{ \frac{ t-25 }{ t } } \mathrm{d} t$
Separate the fraction into $2$ fractions$\frac{ 1 }{ 2 } \times \int{ \frac{ t }{ t }-\frac{ 25 }{ t } } \mathrm{d} t$
Any expression divided by itself equals $1$$\frac{ 1 }{ 2 } \times \int{ 1-\frac{ 25 }{ t } } \mathrm{d} t$
Use the property of integral $\int{ f\left( x \right)\pmg\left( x \right) } \mathrm{d} x=\int{ f\left( x \right) } \mathrm{d} x\pm\int{ g\left( x \right) } \mathrm{d} x$$\frac{ 1 }{ 2 } \times \left( \int{ 1 } \mathrm{d} t-\int{ \frac{ 25 }{ t } } \mathrm{d} t \right)$
Use $\int{ 1 } \mathrm{d} x=x$ to evaluate the integral$\frac{ 1 }{ 2 } \times \left( t-\int{ \frac{ 25 }{ t } } \mathrm{d} t \right)$
Use $\int{ \frac{ a }{ x } } \mathrm{d} x=a \times \ln\left({|x|}\right)$ to evaluate the integral$\frac{ 1 }{ 2 } \times \left( t-25\ln\left({|t|}\right) \right)$
Substitute back $t={x}^{2}+25$$\frac{ 1 }{ 2 } \times \left( {x}^{2}+25-25\ln\left({|{x}^{2}+25|}\right) \right)$
Simplify the expression$\frac{ 1 }{ 2 }{x}^{2}+\frac{ 25 }{ 2 }-\frac{ 25 }{ 2 } \times \ln\left({{x}^{2}+25}\right)$
Add the constant of integration $C \in ℝ$$\begin{array} { l }\frac{ 1 }{ 2 }{x}^{2}-\frac{ 25 }{ 2 } \times \ln\left({{x}^{2}+25}\right)+C,& C \in ℝ\end{array}$