## Thomas' Calculus 13th Edition

$$- \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right) + \frac{1}{2}\left( {{{\tan }^{ - 1}}t} \right) + C$$
\eqalign{ & \int {\frac{{dt}}{{{t^4} + 4{t^2} + 3}}} \cr & {\text{Decompose the integrand into partial fractions.}} \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{1}{{{t^4} + 4{t^2} + 3}} = \frac{1}{{\left( {{t^2} + 3} \right)\left( {{t^2} + 1} \right)}} = \frac{{At + B}}{{{t^2} + 3}} + \frac{{Ct + D}}{{{t^2} + 1}} \cr & \cr & {\text{Multiplying by }}\left( {{t^2} + 3} \right)\left( {{t^2} + 1} \right){\text{, we have}} \cr & 1 = \left( {At + B} \right)\left( {{t^2} + 1} \right) + \left( {Ct + D} \right)\left( {{t^2} + 3} \right)\,\,\,\,\,\left( {\bf{1}} \right) \cr & 1 = A{t^3} + At + B{t^2} + B + C{t^3} + 3Ct + D{t^2} + 3D \cr & 1 = \left( {A{t^3} + C{t^3}} \right) + \left( {B{t^2} + D{t^2}} \right) + \left( {At + 3Ct} \right) + 3D + B \cr & \cr & {\text{If we equate coefficients}}{\text{, we get the system}} \cr & A + C = 0,\,\,\,\,\,\,\,\,\,\,\,\,B + D = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,A + 3C = 0,\,\,\,\,\,3D + B = 1 \cr & {\text{Solving by using a calculator, we get}} \cr & A = 0,\,\,B = - \frac{1}{2},\,\,\,C = 0,\,\,\,\,D = \frac{1}{2} \cr & \cr & {\text{The decomposition of the integrand is}} \cr & \frac{1}{{{t^4} + 4{t^2} + 3}} = \frac{{ - 1/2}}{{{t^2} + 3}} + \frac{{1/2}}{{{t^2} + 1}} \cr & \cr & \int {\frac{{dt}}{{{t^4} + 4{t^2} + 3}}} = - \frac{1}{2}\int {\frac{1}{{{t^2} + 3}}} dt + \frac{1}{2}\int {\frac{1}{{{t^2} + 1}}} dt \cr & \cr & {\text{Integrating gives}} \cr & \int {\frac{{dt}}{{{t^4} + 4{t^2} + 3}}} = - \frac{1}{2}\left( {\frac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right)} \right) + \frac{1}{2}\left( {{{\tan }^{ - 1}}t} \right) + C \cr & \int {\frac{{dt}}{{{t^4} + 4{t^2} + 3}}} = - \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right) + \frac{1}{2}\left( {{{\tan }^{ - 1}}t} \right) + C \cr}