Integration technique

Evaluate





The radical term in the denominator is a hint for trigonometric substitution. However, this problem requires first a clever factoring manipulation and algebraic substitution before trigonometric substitution can be employed.

Let





Another substitution












From

the illustration below follows.









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Partial fractions

I already finished the contents of my workbook X and Why: Integration Techniques. I am now in the reproduction stage. While waiting for the finished product, let me present some of its contents.

Solve the following:
a)



b)



The problems look easy. Indeed, they are. So, how do we solve these?

For me, the basic step is to examine the function. There are several integration techniques but in these problems, it should be obvious that trigonometric integration, trigonometric substitution and integration by parts are not applicable. Why? The answer lies in the nature of the function in the given problems.

The power formula will not work because we cannot split these expressions to produce a function matching its derivative. Algebraic substitution will not work because the expressions are already simplified. The other option, which turns out to be applicable, is integration using partial fractions.

For a, note the like powers for numerator and denominator, so divide first. The first term in the answer should be obvious.



For b, there are non-repeating linear factors.





Easy.


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Reference: [4]

Death

Old mathematicians never die; they just lose some of their functions.

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source: [7]