Eric had a good answer, but I'll give my approaches since it's not about whether they know what to do, but whether they can figure it out without someone telling them what to do.

I prefer to start from the fundamentals to avoid any mistakes. So first off, the trig substitution trio:

#(a^2 + x^2)# reminds me of #1 + tan^2theta = sec^2theta#, and I would use #x = asectheta#.

#(a^2 - x^2)# reminds me of #1 - cos^2theta = sin^2theta#, and I would use #x = asintheta#.

#(x^2 - a^2)# reminds me of #sec^2theta - 1 = tan^2theta#, and I would use #x = atantheta#.

Now let's look at the problems.

**a)**

With this problem, I see the #sqrt(1-9x^2)# and I think about trig substitution. If you see something like #sqrt(pmx^2pma^2)#, consider trig substitution. If you don't see a square root, don't go straight to trig substitution. Think about if you could have done something else, like u-substitution or partial fraction decomposition (hopefully not that).

In this case, we can modify the terms to give a coefficient of 1 on #x^2#. Then we get:

#int 1/(sqrt(9)sqrt(1/9-x^2))dx = 1/3int 1/(sqrt(1/9-x^2))dx#

So, #a = 1/3#. At this point I usually consider what trig identity this reminds me of. When I see #a^2 - x^2#, if I replace #x# with a trig term, I'm drawn to #sin^2theta + cos^2theta = 1#, because #cos^2theta = 1-sin^2theta#. So, let's see what this reduces to.

Let #x = asintheta = 1/3sintheta#. Then that means #dx = 1/3costheta##d##theta#. Now we can substitute.

#1/3int 1/(sqrt(1/9-x^2))dx = 1/3int 1/(sqrt(1/9-1/9sin^2theta))1/3costheta#

#= 1/3int costheta/(cancel(3)cancel(1/3)sqrt(1-sin^2theta))d##theta#

#= 1/3int cancel(costheta/(sqrt(cos^2theta)))d##theta = 1/3intd##theta#

#= 1/3theta + C#

Now, we can work backwards and look at how we said #x = 1/3sintheta#. Therefore, #theta = arcsin(3x)#. So, we get:

#1/3arcsin(3x) + C#

So that one was pretty straightforward, but I don't think it's a very comprehensive practice problem. You didn't even get to draw a right triangle. Ah well.

**b)**

At first I thought about multiplying it out, but then I realized, it was written as #x-2# for a reason.

#int 1/(9 + (x-2)^2)dx = 1/(a^2 + (x-2)^2)#

#a = 3#

#x - 2 = ?#

I look at this, and it reminds me of #1 + tan^2theta = sec^2theta#.

So naturally, we get:

#x - 2 = 3tantheta#

#dx = 3sec^2thetad##theta#

Plug it in, and divide out common factors:

#int 1/(9 + 9tan^2theta)3sec^2thetad##theta#

#= int sec^2theta/(3 + 3tan^2theta)d##theta#

Then, WOAH, what's this? Another substitution we can do? Yeah... that's right. We can do a u-substitution on this. But let's not, because we can use identities.

#= int cancel(sec^2theta)/(3(cancel(sec^2theta)))d##theta#

#= 1/3theta + C#

If you recall, #x-2 = 3tantheta#, so #theta = arctan((x-2)/3)#. So, we get:

#1/3arctan((x-2)/3) + C#

**c)**

#int (x-3)/(x^2+1)dx#

Already this looks interesting. What might trip people up is the #x-3#, because it is not a common factor of #x^2 + 1#. However, you can separate the fraction.

#int (x-3)/(x^2+1)dx = int x/(x^2+1)dx - int 3/(x^2+1)dx#

You may notice that there is no need for trig substitution here. There isn't. You now have a #1/(1+x^2)# integral in the second one, which if you recall, is #arctanx + C#. The first one gives you #x^2# and #xdx#, which resembles a u-substitution problem.

Let:

#u = x^2 + 1#

#du = 2xdx#

#xdx = 1/2du#

#int x/(x^2+1)dx - int 3/(x^2+1)dx = 1/2int 1/udu - 3int 1/(x^2+1)dx#

That's about it. Just integrate and get your #ln|u|# and #arctanx#:

#= 1/2ln|x^2+1| - 3arctanx + C#

Eric, note that you wrote:

#x = tantheta#

which is fine, because #a = 1#. But you also wrote:

#2dx = sec^2thetad##theta#

But #dx = sec^2thetad##theta#, so #2dx = 2sec^2thetad##theta#.

So, the answer would not contain #3/2arctanx#, but #3arctanx#. :) (Also, #int1/udu = ln|u| + C#, rather than #lnu + C#, but that's a minor point since #x^2 + 1 > 0# anyways)