i have reached a chapter in my pure mathematics text that introduces u-substitution as a method for solving integrals where in general if f(x) = dy/dx, an integral with respect to x may be transformed into an integral with respect to a related variable u such that: ∫ f(x) dx = ∫ [f(x) dx/du]du = y.(adsbygoogle = window.adsbygoogle || []).push({});

A number of examples are carefully illustrated to demonstrate this technique. There is one problem in particular whereby the worked solution is as follows;

Find; ∫ x√(3x-1)dx

Solution;

let: u = √(3x-1), thus.. dx/du = 2u/3 and let: x=1/3(u^2 +1).

→ ∫ [x√(3x-1) dx/du]du = ∫ [1/3(u^2+1)u 2u/3]du

= ∫ [2/9(u^4) + 2/9(u^2)]du

The problem i have here is that when i separate the variables and integrate the [dx/du = 2u/3] i get x=1/3(u^2) + c, and not x=1/3(u^2 +1) as shown in the book. i understand the further simplification from here and i can see that setting the constant of integration to one (c=1) to give x=1/3(u^2 +1) is perhaps some kind of trick that allows one to get the second term in terms of the derivative of x, w.r.t u? but i'm unable to figure out how the intuition for this works out, after all could that constant not be any real number?

I have not seen this kind of substitution before. i only solved for and substituted for u and du so far. Can anyone explain to me what kind of u substitution this is?

I am new to the forum as an official member and would like to take opportunity to say thank you to all the members and contributors to this site, i have read many informative and interesting problems and discussions over the years and as a resource has been astonishingly useful to me. Thanks again! :)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Having some trouble understanding a u-substitution

**Physics Forums | Science Articles, Homework Help, Discussion**