Integration by substitution (often referred to as u-substitution) is quite hard to understand. Most people just follow their proven recipe when performing this kind of integration. In contrast to that doing, I want to illustrate why the steps of u-integration are necessary. Therefore we will develop the idea slowly by giving simple examples which illustrate what works and what doesn’t.

What integration by substitution gives you

Say, you wanted to tackle an integral like the following:

\displaystyle \int x \sqrt{x-1}\, dx

You cannot integrate because the radical prevents you from doing so: Neither are you able to integrate the whole product nor does rearranging the term helps you. With x-1=u however the term in the radical disappears. So substitue x=u+1. Now you have to deal with the expression

\displaystyle \int (u+1) \times \sqrt{(u+1)-1}\, dx=\int (u+1) \times \sqrt{u}\, dx=\int \sqrt{u^{3}}+\sqrt{u}\, dx

This is obviously easy to integrate. We obtain the solution

\displaystyle \int \sqrt{u^{3}}+\sqrt{u}\, du=\frac{2u^{\frac{5}{2}}}{5}+\frac{2u^{\frac{3}{2}}}{3}+c

in which we can substitute back u=x-1  to receive the solution of the integral from the beginning of the section:

\displaystyle \int x \times \sqrt{x-1}\, dx=\frac{2(x-1)^{\frac{5}{2}}}{5}+\frac{2(x-1)^{\frac{3}{2}}}{3}+c

If you read about u-integration before, you may feel the need of some steps which I left out here (like substituting dx). Nevertheless this example works. Take a look at the following picture. It clearly shows how substituting shifted the graph horizontally.

Comparing the original to the substituted function

As you already have guessed, the integrals of both functions are identical. They only differ in location, but we fixed this by substituting backwars in the end. The main goal of this section however was to show you that substitution allows you get rid of inconvenient terms.

Why it’s not always that easy

Let’s try another example. You already know that \int 2x^{2}\, dx results in (2/3)x^{3}+c. To receive some interactive experience, you may want to solve the integral

\displaystyle \int x\times 2x\, dx

by substituting x=u/2 . It gives

\displaystyle \int \frac{u}{2}\times 2\frac{u}{2}\, du=\frac{u^{3}}{6}+c=\frac{(2x)^{3}}{6}+c=\frac{4}{3}x^{3}+c\neq\frac{1}{3}x^{3}+c

The result is double as high as expected, but why? From the following graphic you can deduce that the transform due to substitution changed limits and value of the integral. Say you wanted to find the definite integral within the limits x=0 and x=0.5. This range corresponds to the range u=0 to u=1. Comparing both functions indeed shows that the areas below the graphs differ.

Comparing the original to the substituted function

It is not always enough to only substitute x! The next section will explain you why.

Substituting dx – The missing factor!

Actually we omitted the substitution of dx up to this point. Instead we just replaced dx  by du which is inaccurate and works only in some special cases of integration by substitution. The picture below illustrates some coherences.

As a first step, we differentiate our integral whose integrand we name f'(x). It becomes evident that substituting x  by x(u) cannot work if you only do so on the right-hand side of the equation at line 2. In line 3 we switch from differentiating in terms of x to differentiating in terms of u. Since x is now an inner function in terms of u, we usually need to apply the chain rule. The chain rule in turn involves multiplying by x'(u), but this factor is missing yet in our calculation!

Consequences of u-substitution

When taking a look at the example from the previous section, you will notice that x'(u)=1/2 and hence

\displaystyle \int \frac{u}{2}\times 2\frac{u}{2} \times \frac{1}{2}\, du=\frac{2}{3}x^{3}+c

which is the correct solution! You may have seen shortcuts based on the fact that dx/du=x'(u)  or du/dx=u'(x) (the latter will require substituting x  again) . Using these facts you can directly substitute into the integral, which is not the best way regarding mathematical correctness – but it is handy!

\displaystyle dx=x'(u)\, du=1/u'(x(u))\, du=\frac{1}{2}\, du

\displaystyle \int \frac{u}{2}\times 2\frac{u}{2}\, dx=\int \frac{u}{2}\times u \times \frac{1}{2}\, du

A final example

Integration by substitution often doesn’t work solely. I will show you an example which will involve integration by parts as well. Further, the example will illustrate how to work with limits directly instead of performing backwards-substitution in the end. Here we go!

\displaystyle A=\int\limits_{0}^{1} e^{\sqrt{x+1}}\, dx

We simplify using u(x)=\sqrt{x+1}  to transform the limits and its inverse x(u)=u^{2}-1  to transform the integrand.

\displaystyle A=\int\limits_{\sqrt{0+1}}^{\sqrt{1+1}} e^{\sqrt{(u^{2}-1)+1}}\, dx

Simplifying and substituting dx=x'(u)\, du=2u\, du gives:

\displaystyle A=\int\limits_{1}^{\sqrt{2}} e^{u}\times 2u \, du

Integration by parts comes in here:

\displaystyle A=\left[e^{u}\, 2u \right]_{1}^{\sqrt{2}}- \int\limits_{1}^{\sqrt{2}} e^{u}\times 2 \, du=\left[2u\, e^{u} - 2e^{u}\right]_{1}^{\sqrt{2}}

\displaystyle A=2e^{\sqrt{2}} (\sqrt{2}-1)-2e^{\sqrt{1}} (\sqrt{1}-1)=2e^{\sqrt{2}} (\sqrt{2}-1)

You could also have substituted backwards as usual using the fact that u=\sqrt{x+1}  to obtain the indefinite integral:

\displaystyle 2u\, e^{u}-2e^{u}=2e^{u}(u-1)=2e^{\sqrt{x+1}}(\sqrt{x+1}-1)+c

I hope you now agree that integration by substitution (alias u-substitution) is an easy, logical and helpful thing 🙂