# Integration techniques | U substitution – part 2

U-substitution is the most common technique used in integration. It can happen, however, that it doesn’t work out, no matter what you try to substitute. You may think of using another technique, like integration by parts, but it’s not always necessary. You realise u-sub is not working when you still see one or more x’s. For example:

$\int&space;\ln(x)dx$ $u=\ln(x)\;&space;\;&space;\;&space;\;&space;du=\frac{1}{x}dx\rightarrow&space;dx=xdu$ $\int&space;u\,&space;x\,&space;du$

As you can see, we have an x that we don’t want. At this point you may think about integration by parts, and yes, that’s actually the way to go. The result is x(ln(x)-1), or xln(x)-x. But it is possible to keep going with u-sub. Think of x in terms of u. We let u = ln(x), which means that x = e^u. So we can write as follows:

$\int&space;u\,&space;e^u\,&space;du$

As you can see, everything is in the u-world now. At this point you can perform integration by parts, undo the u-sub and you’ll get the same result, xln(x)-x. This was just an example, to show you that the x you see can be written in terms of u. If this method doesn’t work, you can use the third type of u substitution. You can use this when a function of x, like 1-x, is inside another function, like the square root. Let’s use this example:

$\int&space;\sqrt{1-x}dx$

You can, and should, use the first method, which is by letting u be equal to 1-x but again, to show you this other way of proceding, I’m not going to do that. Instead, let u = sqrt(1-x).

$u=\sqrt{1-x}\,&space;\,&space;\rightarrow&space;\,&space;u^2=1-x$ $2udu=-dx\rightarrow&space;dx=-2udu$ $\int&space;u(-2udu)=-2\int&space;u^2du=-\frac{2}{3}u^3\rightarrow&space;-\frac{2}{3}\left&space;(&space;\sqrt{1-x}&space;\right&space;)^3+C$

The fourth method constists in extracting the x from u = f(x) and take the derivative on both sides.

Example:

$\int&space;x^2dx$ $u=x^2\rightarrow&space;x=\sqrt{u}\;&space;\;&space;\;&space;\;&space;\;&space;dx=\frac{1}{2\sqrt{u}}du$ $\int&space;u\frac{1}{2\sqrt{u}}du=\frac{1}{2}\int&space;u^{\frac{1}{2}}du=\frac{1}{3}u^{\frac{3}{2}}\rightarrow&space;\frac{1}{3}\left&space;(&space;x^2&space;\right&space;)^{\frac{3}{2}}=\frac{1}{3}x^3+C$

This is a total nonsense, since we just need to apply the power rule to x^2. It was just to show you what you can do when none of the other methods work.

The next integration technique is trigonometric substitution, which I will talk about in another post. Next post is going to be about integration of a function of x with respect to another function of x. Stay tuned!

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