Evaluating integrals might seem challenging sometimes, especially when we encounter very complex looking expression or sometimes even when we encounter simple looking integrals such as ∫sec(x) dx, which turns out to involve some cleverness when evaluating it. Being “good” at integration might be “innate” for some people, but often a lot of practice is the key. When practicing you might notice that you get an “instinct” on how to solve certain types of integral even at first sight and how to apply certain methods, like when and how to apply  integration by susbtitution method.  Here are some of the tips I got from my hours of practice in how to make a good substitution:

• When encountering a binomio in the form 3: (ax+b)^n. Solving an expresion like this will take a lot of time and effort. In this case, a good way to proceed is:

∫(3x-34)^32 dx

u= 3x-34 ; let the binomio = u, then du=6 dx

∫(3x-34)^32 dx = ∫(u)^32 du/6 = (1/6) ∫u^32 du = (1/6) [u^(32+1) /(32+1)]

• When encountering a fraction in the form: x^(n-1)/x^n +k  or x/x+k (k is a constant). How to proceed:

∫ x^4 / (x^5-3x²) dx=∫ x^4 / [ x² (x³-3) ] dx = ∫ x² / (x³-3) dx

u=x³-3 ; let u= the denominator, the du=3x²dx → dx=du/3x²

∫ x² / (x³-3) dx = ∫ x² / u ( du / 3x² ) = ∫ 1/u ( du/3 ) = 1/3  ∫ 1/u du = 1/3 log |u|

• When encountering log(x). This expression can appear in different ways and you might not choose to evaluate the integral always the same, but a good way to start is:

∫14x² / [x³ log²(x)] dx = ∫14/xlog²(x) dx = 14 ∫ 1 / xlog²(x) dx

u=log(x) ; choose u=log(x), then du=1/x dx → dx=x du

14 ∫ 1 / xlog²(x) dx = 14 ∫ (x du)/(xu²)  = 14∫ 1/u² du = 14 u‾¹/-1 = 14 (-1/u)