Solving difficult integrals with substitution

Integration by substitution is a very powerful tool for evaluating integrals where the answer may not seem as obvious at first, but under appropriate variable transformations, we can obtain simple and elegant solutions. In this article, I will explore 2 interesting examples of substitution which caught my eye, and which I think illustrate the power of this method.

Integral 1

This integral seems tame at first, because we know that the integral of an exponential function is the same function again times some constant, but in this case, we have a 1 added to exp(ax) in the denominator, which means finding the anti-derivative of this function (and hence evaluating the definite integral) is not as straight-forward. Luckily, we can make a variable substitution which allows us to simplify things. Suppose that since we know how to integrate exp(ax), we make the substitution

and therefore, the derivative of u with respect to x is

Next, we evaluate the new limits in the domain of u. The limits in x are 0 and infinity, therefore:

Therefore, the new integral becomes:

This last expression can be evaluated by partial fraction decomposition, that is: we let