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. 

You did nothing wrong. Here are two alternate approaches:

$$\int\frac{dt}{t^3\sqrt{t^2-1}}=\int\frac{t}{t^4\sqrt{t^2-1}}~dt=\frac12\int\frac{d\big(t^2-1\big)}{\Big[\big(t^2-1\big)+1\Big]^2\sqrt{t^2-1}}=\frac12\int\frac{du}{\big(u+1\big)^2\sqrt u}$$
$$=\frac12\int\frac{d\big(v^2\big)}{\big(v^2+1\big)^2~v}=\int\frac{dv}{\big(v^2+1\big)^2}$$

Or, substituting $t=\cosh x$, and using $\cosh'x=\sqrt{\cosh^2x-1}=\sinh x$, we have

$$\int\frac{\sinh x}{\cosh^3x~\sinh x}~dx=\int\frac{\cosh x}{\cosh^4x}~dx=\int\frac{d\big(\sinh x\big)}{\big(1+\sinh^2x\big)^2}=\int\frac{dv}{\big(v^2+1\big)^2}$$

Using the fact that $~\dfrac1{\big(v^2+1\big)^2}=-\bigg[\dfrac{d}{da}\bigg(\dfrac1{v^2+a}\bigg)\bigg]_{a=1}$ and $~\displaystyle\int\frac{dv}{v^2+a}=\frac1{\sqrt a}~\arctan\frac v{\sqrt a}$
we soon arrive at the desired conclusion, $I=-\bigg[\dfrac{d}{da}~\bigg(\dfrac1{\sqrt a}~\arctan\dfrac v{\sqrt a}\bigg)\bigg]_{a=1}=~\dfrac{\arctan v}2~+$
$+~\dfrac12\cdot\dfrac{v}{v^2+1},~$ which, when evaluated between $v=\sqrt3$ and $v=1$, yields $~\dfrac{\pi+3\sqrt3-6}{12}~.$
