Evaluate the Integral $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$ $\int^2_{\sqrt{2}}\frac{1}{t^3\sqrt{t^2-1}}dt$
I believe I've done everything right; however, my answer does not resemble the answer in the book. I think it has something to do with my Algebra. Please tell me what I am doing wrong. 

Your integral is well calculated.
Computing the last part for a prominent answer, we get
$$\frac{1}{2}[1+\frac{1}{2}\sin 2\theta]^{\frac{\pi}{3}}_{\frac{\pi}{4}}$$
$$=\frac{1}{2}[(\frac{\pi}{3}-\frac{\pi}{4})+\frac{1}{2}(\sin (\frac{2\pi}{3})-\sin (\frac{2\pi}{4}))]$$
$$=\frac{1}{2}[\frac{\pi}{12}+\frac{1}{2}(\frac{\sqrt{3}}{2}-1)]$$
$$=\frac{\pi}{24}+\frac{1}{8}(\sqrt{3}-2)$$
Does it match the answer?
