Problem: Evaluate the integral

Solution: We started to solve this problem in this note as an example of substitution, we prepared it like this:

Why did we chose to do so? The root was clearly troublesome, so getting rid of it by substitution seemed like a good idea. Whether it will be possible or not depended on us being able to express dx solely in terms of y. We succeeded, thus we can pass to a new integral.

Were there any alternatives? When we go through "boxes" for various types, we see that the given integral fits only the box "integrals with roots". There the recommendation is to use the indirect substitution x = y2, but that is exactly what we did here.

The new integral clearly belongs to the box "rational function", so we use the appropriate procedure. Just to be sure that we do not work unnecessarily we check whether it is possible to cancel something in the fraction, but it does not look like it, so we start standard steps. Since the degree in the numerator is higher, we first apply the long division and then decompose the resulting remainder into partial fractions:

Since there are only linear factors, we easily obtain the unknown constants using the cover-up trick and the decomposition is done.

Since we know that a linear substitution always works, we see that both fractions are essentially of the form 1/y, which means an elementary integral. Also the polynomial y − 2 gives an elementary integral, so we can finish the calculation. If you are not sure about integrating the partial fractions, play it safe and apply substitutions, say, s = y − 1 and t = y + 3; see here. So we have

By the way, if the question asked for the indefinite integral (or if we decided to first find an antiderivative and then plug in the limits), we would get

We could put the logarithms together, but it would lead to a fifteenth power, which would not look any better than what we already have, so we leave the answer alone. The result is valid on all intervals of non-negative numbers not including x = 1. When calculating definite integrals, it is important to check that we integrate over intervals where an antiderivative exists, which is definitely true for the given integral.


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