So I've been reading about integration these past few days by myself, and my book has a table where it shows me how to integrate some elementary functions (like $x$ to the power of $n$, or $1/x$, or $e^x$, $e^{Kx}$, $\cos x$, $\sin x$, $\ln(x)$), and I'm pretty cool with those, but I'm having a bit of trouble when it gets a more complex:

For example: $$\int\frac1{2\sqrt x}\,dx = {?}$$

I mean, do I simply just follow the rule that says $\int 1/x\,dx = \ln(x)$, and say $\int1/(2\sqrt x)\,dx = \ln(2\sqrt x)$? Doesn't seem right to me.

Another example from the textbook:

$$\int\frac{\ln(x)}{\sqrt x}\,dx$$

No idea what to do.

It's primarily fractions that I don't get. I know how to solve $k \cdot f(x)$ (integrate $f(x)$, keep the constant), $f(x)+g(x)$ (integrate separately and add), $f(x)-g(x)$ (integrate separately and subtract), as well as the methods of partial integration and integration by substitution.

But how I deal with fractions, especially those without clean numbers like above?

  • $\begingroup$ $\frac{d \sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}}$ $\endgroup$ – user76568 Mar 6 '14 at 17:49
  • $\begingroup$ ${1\over 2 \sqrt x} = {x^{-1/2} \over 2 }$. $\endgroup$ – Karolis Juodelė Mar 6 '14 at 17:50

There are many techniques for integration, which your book will discuss. For $\int \frac 1{2\sqrt x}dx$ (please supply parentheses to show the square root is in the denominator-it could be read as $\frac 12 \sqrt x$), note that $\sqrt x=x^{1/2}$ so we have a form you know $$\frac 1{2\sqrt x}dx=\int \frac 12 x^{-\frac 12}=\frac 12\cdot \frac {x^ {\frac 12}}{\frac 12}+C=\sqrt x+C$$ There is not an algorithm like differentiation. There are a number of techniques, each of which works some of the time. Sometimes, none of them work.

  • $\begingroup$ Your book will discuss ??? Maybe. Or, if that book doesn't discuss it, wait until you get to a book that does before asking about it... $\endgroup$ – GEdgar Mar 6 '14 at 20:21

Two hints:

  1. Rewrite $\frac{1}{\sqrt{x}} = x^{-1/2}$ and use the integration rule $\int x^n dx = \frac{1}{n+1} x^{n+1}$, which holds not only for natural numbers $n$, but also for fractional numbers $n \neq -1$.
  2. For $\int \ln x / \sqrt{x} dx$ set $u=\ln x$ and $v' = 1/\sqrt{x} = x^{-1/2}$, then use partial integration.

Unlike differentiation rules, it needs a lot of training to "see" which rule applies. If you aren't lucky, maybe you have to think harder and try some of your learned techniques. If you have access to a good library, there are some books containing lists of integrals, e.g. from the Eastern part of Europe:

  • Gradstein/Ryshik
  • Bronstein/Semendyayev
  • Prudnikov/Brychkov

However, most of them don't give you hints, how they achieved their results...


Regarding $\frac{\ln{x}}{\sqrt{x}}$, using integration by parts: $$\int \frac{\ln{x}}{\sqrt{x}} dx = \int 2\ln{x} \frac{d\sqrt{x}}{dx}dx = 2 \ln{x} \sqrt{x} - \int\frac{2}{x} \sqrt{x} dx = 2 \ln{x} \sqrt{x} - \int\frac{2}{\sqrt{x}} dx$$

Now you can use Ross Millikan's answer.


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