Problem: Evaluate the integral

Solution: We can get to the solution of this problem in two ways. Usually it pays to consider substitution as the first option and here obviously the root causes trouble. Would it work if we denote is by y? The usual criterion of success is to differentiate the considered expression and check whether we can see this derivative next to dx. That is not true here, but the more experienced already know that substitutions with roots can be often fixed to work. We will try it.

So it worked. Before we move on we will explore another starting point. This integral obviously belongs to the box "integrals with roots" and there they recommend to use mixed substitution to get rid of the root, in this case x + 2 = y2. This is the same substitution we tried above. The third way to get to it we will show just for the sake of completeness, since it is not exactly very natural. We can imagine that the expression x + 2 is in fact the ratio (x + 2)/1 and then check on the box "ratio of linear functions", where they recommend our substitution.

Back to our problem. We obtained an integral of a rational function. We have to start by the long division, the remainder will be some constant divided by a linear expression, which easily leads (after a linear substitution) to a logarithm:

By the way, if we needed the indefinite integral, after back substitution we would have (try it)

If you are looking at the answer you got and wondering where did the two go, we hid it in the constant C. The domain of the integral is determined by two conditions. First, the expression under the root cannot be negative, which leads to a closed half-interval. Second, the denominator cannot be equal to zero, which makes a "hole" in this interval across which we cannot integrate. Fortunately, in our problem we integrate over an interval that is away from this hole.

There is another substitution one can try: to denote the whole denominator. Such an ambitious substitution usually end up badly and it is better to be modest and play it safe, but let's try it.

So this would also work, perhaps it is even easier than our first solution (we were unusually lucky), but the result looks worse. Try to multiply out those powers to check that it is the same as our first result.


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