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My textbook asked me to evaluate the integral $\int_1^5\left(\frac{x}{\sqrt{2x-1}}\right)$ using u-substitution. I rewrote the integrand as $x*(2x-1)^{-\frac{1}{2}}$, but I soon became stuck because I couldn't settle on what was $g'(x), g(x),$ or $F(x)$ (I didn't know how to define u or du). I checked the book (this was an example problem) and it said to set u = $\sqrt{2x-1}$. How did it come to this conclusion? Thanks for your help.

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  • $\begingroup$ What else would you substitute in this situation? $u=2x-1$ should also work. $\endgroup$ – Peter Foreman Jun 15 at 0:20
  • $\begingroup$ I'm just asking how it got to this solution. To clarify, how did it identify the composed functions? $\endgroup$ – N. Bar Jun 15 at 0:24
  • $\begingroup$ Very often the idea is to get rid of the radical. So, ... $\endgroup$ – Claude Leibovici Jun 15 at 3:32
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They are trying to get you to use the trick that setting $$u=\sqrt{2x-1}\implies\frac{u^2+1}{2}=x\implies u \space du=dx$$ Then $$I=\int_{x=1}^{x=5}\frac{x}{\sqrt{2x-1}}dx=\int_{u=1}^{u=3}\frac{\left(\frac{u^2+1}{2}\right)}{u}\cdot u\space du=\frac{1}{2}\int_1^3(u^2+1)\space du$$

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In general, it can be difficult to identify what substitutions to perform, and even harder in most cases to identify them in terms of a sort of function composition. For some cases, composition is simple to notice: $$\int 2x e^{x^2}dx=\int e^udu=e^{x^2}+C$$ But in more complicated examples, $u$-substitutions can be difficult to immediately see. It is often helpful, instead of considering function composition, to see what parts of the integrand may be simplified or cancelled through a particular substitution. In your integral, substituting $u=\sqrt{2x-1}\implies du=\frac{dx}{\sqrt{2x-1}}=\frac{dx}{u}$ is helpful - the substitution transforms the $x$ in the numerator into a quadratic in $u$ (since $u=\sqrt{2x-1}\implies x=\frac{u^2+1}{2}$), the denominator simply becomes $u$, and we multiply by $u$ when we change from $dx$ to $u\,du$. The integral becomes: $$\int_1^5\frac{x}{\sqrt{2x-1}}\,dx=\frac{1}{2}\int_1^3\frac{u^2+1}{u}\,u\,du=\frac{1}{2}\int_1^3\left(u^2+1\right)du\text,$$ which we can integrate very easily. In this case, it may not have been obvious that this integral was an example of function composition, but it was much easier to play around with different substitutions and notice that this particular one led to some very convenient simplifications and cancellations.

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Alternatively: $$\int_1^5\left(\frac{x}{\sqrt{2x-1}}\right) \ dx=\frac12\int_1^5\left(\frac{2x-1+1}{\sqrt{2x-1}}\right) \ dx=\\ \frac12\int_1^5\left[\sqrt{2x-1}+ \frac1{\sqrt{2x-1}} \right] \ dx=\\ \frac12\left[\frac13(2x-1)^{3/2}+(2x-1)^{1/2}\right]\bigg{|}_1^5=\\ \frac12\left[(9+3)-(\frac13+1)\right]=\frac{16}{3}.$$

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