Answer
$4\ln \sqrt 3 + \frac{\pi }{{12}} - \frac{1}{2}\ln 2$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt 3 } {\frac{{3{t^2} + t + 4}}{{{t^3} + t}}dt} \cr
& {\text{Decompose the integrand }}\frac{{3{t^2} + t + 4}}{{{t^3} + t}}{\text{ into partial fractions}} \cr
& \frac{{3{t^2} + t + 4}}{{{t^3} + t}} = \frac{{3{t^2} + t + 4}}{{t\left( {{t^2} + 1} \right)}} = \frac{A}{t} + \frac{{Bt + C}}{{{t^2} + 1}} \cr
& {\text{multiply by }}t\left( {{t^2} + 1} \right){\text{ and simplify}} \cr
& 3{t^2} + t + 4 = A\left( {{t^2} + 1} \right) + \left( {Bt + C} \right)t\,\,\,\, \cr
& {\text{Expand the equation }} \cr
& 3{t^2} + t + 4 = A{t^2} + A + B{t^2} + Ct\, \cr
& {\text{Group terms}} \cr
& 3{t^2} + t + 4 = \left( {A{t^2} + B{t^2}} \right) + Ct\, + A \cr
& {\text{Equating coefficients}} \cr
& A = 4,\,\,\,\,A + B = 3\,\,\,\, \to \,\,\,\,B = - 1,\,\,\,\,\,\,C = 1 \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{3{t^2} + t + 4}}{{t\left( {{t^2} + 1} \right)}} = \frac{4}{t} + \frac{{ - t + 1}}{{{t^2} + 1}} \cr
& \frac{{3{t^2} + t + 4}}{{t\left( {{t^2} + 1} \right)}} = \frac{4}{t} - \frac{t}{{{t^2} + 1}} + \frac{1}{{{t^2} + 1}} \cr
& \int_1^{\sqrt 3 } {\frac{{3{t^2} + t + 4}}{{{t^3} + t}}dt} = \int_1^{\sqrt 3 } {\left( {\frac{4}{t} - \frac{t}{{{t^2} + 1}} + \frac{1}{{{t^2} + 1}}} \right)dt} \cr
& {\text{Integrating}} \cr
& = \left( {4\ln \left| t \right| - \frac{1}{2}\ln \left( {{t^2} + 1} \right) + \arctan t} \right)_1^{\sqrt 3 } \cr
& = \left( {4\ln \left| {\sqrt 3 } \right| - \frac{1}{2}\ln \left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right) + \arctan \sqrt 3 } \right) - \left( {4\ln \left| 1 \right| - \frac{1}{2}\ln \left( {{{\left( 1 \right)}^2} + 1} \right) + \arctan 1} \right) \cr
& = 4\ln \sqrt 3 - \frac{1}{2}\ln 4 + \frac{\pi }{3} + \frac{1}{2}\ln 2 - \frac{\pi }{4} \cr
& = 4\ln \sqrt 3 + 3\ln 2 + \frac{\pi }{{12}} - \frac{3}{2}\ln 2 \cr
& = 4\ln \sqrt 3 + \frac{\pi }{{12}} - \frac{1}{2}\ln 2 \cr} $$
