Answer
$(\dfrac{1}{6}) \ln |t^2-2|-(\dfrac{1}{6}) \ln |t^2+1|+c$
Work Step by Step
Apply the integration by parts formula as follows: $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Re-write the integral into partial fractions.
$\int \dfrac{1}{t^4-t^2-2}dt=\int [(\dfrac{1}{3}) \dfrac{t}{t^2-2}-(\dfrac{1}{3}) \dfrac{t}{t^2+1}] dt$
Suppose $u=(t^2-2) \implies du=(2t) \space dt\\ a=t^2+1 \implies da=(2t) \space dt$
Simplify.
$ \int [(\dfrac{1}{3}) \dfrac{t}{t^2-2}-(\dfrac{1}{3}) \dfrac{t}{t^2+1}] dt \\=\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$
