Evaluate $\displaystyle \int \frac{dx}{x(x-1)^2} $
This integral should be done by Partial Fractions.
The decomposition you want is $\displaystyle \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$
The function you need to integrate is $\displaystyle \frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}$
Once you're done, use a triangle to switch this back to being in terms of $x$.
$\displaystyle \ln{|x|} - \ln{|x-1|} - \frac{1}{x-1} + C$
Evaluate $\displaystyle \int \frac{x}{(x^2 - 1)^{3/2}}\ dx $
This is just a substitution problem.
Set $u = x^2 - 1$, so that $du = 2x\ dx$.
$\displaystyle \frac{1}{\sqrt{x^2 - 1}} + C$
Compute $\displaystyle \int \frac{x^3}{(x^2 - 1)^{3/2}}\ dx$
This can be done with either trigonometric substitution or normal substitution.
For trigonometric substitution, set $x = sec(\theta)$. For normal substitution, set $u = x^2 - 1$ and manipulate whatever is left in the numerator to be in terms of $u$.
$\displaystyle \sqrt{x^2 - 1} - \frac{1}{\sqrt{x^2 - 1}} + C$
Compute $\displaystyle \int \frac{x^4 + 1}{x^2 + 1} \ dx$
There are a few ways to handle this integral here. The easiest way might be long division into using known integral formulas.
Long division gives that $\displaystyle \frac{x^4 + 1}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1}$
$\displaystyle \frac{x^3}{3} - x + 2 \tan^{-1}(x) + C$
Compute $\displaystyle \int \tan{x} \sec^{5/4}{x}\ dx$
This is related to both trigonometric integrals and substitution.
What choice could I make for $u$ and $du$ here to make this easy? Look for $du$.
Take $du = \sec{x} \tan{x}$, which comes from $u = \sec{x}$.
This integral then becomes $\displaystyle \int u^{1/4}\ du$
$\displaystyle \frac{4}{5} \sec^{5/4}(x)+ C$
Compute $\displaystyle \int e^x \sqrt{e^{2x} - 1}\ dx $. Take your time on this one. It's going to need many different techniques.
Your best bet here is to start with a substitution, and see where that gets you.
Try $u = e^x$ and see what is left, which should leave you with needing to integrate $\sqrt{u^2 - 1}$, which is a trigonometric substitution.
After this, you'll get to integrating $\sec{\theta} \tan^{2}(\theta)$, which requires trigonometric integrals.
This is the same as integrating $\sec^3(\theta) - \sec(\theta)$, which needs reduction, and then you need to convert back to $x$.
$\displaystyle \frac{1}{2}(e^x \sqrt{e^{2x} - 1} - \frac{1}{2} \ln|e^x + \sqrt{e^{2x} - 1}| + C$
Compute $\displaystyle \int x \sec^{-1}{x}\ dx$.
Compute $\displaystyle \int \sqrt{x^2 + 6x}\ dx $