Answer
$(\dfrac{1}{6}) \ln |t^2-2|-(\dfrac{1}{6}) \ln |t^2+1|+c$
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 $\int \dfrac{1}{t^4-t^2-2}dt$
Write the integral into partial fractions as follows: $\int \dfrac{1}{t^4-t^2-2}dt=\int [(\dfrac{1}{3}) \dfrac{t}{t^2-2}-(\dfrac{1}{3}) \dfrac{t}{t^2+1}] dt$
Let us consider $u=(t^2-2) \implies du=(2t) dt$ and $a=t^2+1 \implies da=(2t) dt$
This implies that $\int (\dfrac{1}{6}) \dfrac{du}{u}-\int(\dfrac{1}{6}) \dfrac{da}{a}=(\dfrac{1}{6})\ln |u|-(\dfrac{1}{6})\ln |a|=(\dfrac{1}{6}) \ln |t^2-2|-(\dfrac{1}{6}) \ln |t^2+1|+c$
