Answer
$(\dfrac{1}{2}) \tan^{-1} t-\dfrac{\sqrt 3}{6}\tan^{-1} (\dfrac{t}{\sqrt 3})+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Solve $\int \dfrac{1}{t^4+4t^2+3}dt$
Write the integral into partial fractions as follows: $\int \dfrac{1}{t^4+4t^2+3}dt=\int [(\dfrac{1}{2}) \dfrac{1}{t^2+1}-(\dfrac{1}{2}) \dfrac{1}{t^2+3}] dt$
This implies that $(\dfrac{1}{2}) \tan^{-1} t-\dfrac{1}{2\sqrt 3}\tan^{-1} (\dfrac{t}{\sqrt 3})+c=(\dfrac{1}{2}) \tan^{-1} t-\dfrac{\sqrt 3}{6}\tan^{-1} (\dfrac{t}{\sqrt 3})+c$
