Note To understand this section, you should be familiar with antiderivatives, as discussed in the preceding tutorial (press the button on the sidebar to go there).

The technique of substitution or change of variables is based on the chain rule for derivatives. In the following example, we evaluate the integral

\int 5x(2x^2+1)^{-3} dx.
to illustrate the main steps in changing a variable through substitution. First, we must decide what function to represent as u. Let us follow the advice in Section 6.2 on in or Section 13.2 in : Take u to be an expression that is being raised to a power.

Thus, we will take

u = 2x^2 + 1.

We now follow a fairly mechanical step-by-step procedure:

Step 1:
Calculate the derivative of u, and then solve for "dx."

We usually put this calculation in a little box to separate it from the calculation of the integral.

u = 2x^2 + 1
\frac{du}{dx} = 4x
dx = \frac{1}{4x}du
Step 2:
Substitute the expression for u in the original integral, and also substitute for dx.
\int 5x(2x^2 + 1)^{-3}dx = \int 5xu^{-3}\frac{1}{4x} du
Step 3:
Eliminate the variable x, if it is still present, leaving an integral in u only.

Often, as happens here, all the xs will cancel leaving an expression in u only. (Sometimes, they do not—see below ).

\int 5xu^{-3}\frac{1}{4x} du = \int 5u^{-3}\frac{1}{4} du
Step 4:
Simplify the integrand.

Here, we can sneak the constant 5/4 outside the integral sign.

\int 5u^{-3}\frac{1}{4} du = \frac{5}{4}\int u^{-3} du
Step 5:
Evaluate the simplified integral.

Important: Do not substitute back for u until after this step.

\frac{5}{4}\int u^{-3} du = \frac{5}{4}\frac{u^{-2}}{-2} + C
= -\frac{5u^{-2}}{8} + C
Final Step:
Substitute back for u to obtain the result.
-\frac{5u^{-2}}{8} + C = -\frac{5(x^2 + 1)^{-2}}{8} + C

Thus, \int 5x(2x^2+1)^{-3} dx = -\frac{5(x^2 + 1)^{-2}}{8} + C.

In some of the quizzes below you will have to fill in the answers on your own. Remember to use proper technology/graphing calculator format to input your answers (spaces are ignored).

1. Consider the integral

An appropriate substitution is:

Now substitute in the integral and simplify:

2. Now we look at a more complicated integral:

An appropriate substitution is:

Now substitute in the integral and simplify:

Sometimes, rather than an expression raised to a power, we have a number (such as e) raised to an expression. In such cases, the following advice is often useful:

Take u to be an expression that appears in the exponent.

 

Using the above substitution, we find:

When the xs do not cancel

Q This is all very well, but what do I do if the xs do not cancel?
A Let us go through one such example and see...

Now, we must eliminate any xs that remain, so we do the following:

Go back to the equation that gives u as a function of x, and solve for x.

Now substitute this expression for x in the integral and complete the calculation:

Shortcuts

Notice that the first integral in the above quiz had the form

\int (ax + b)^n dx.
It will be very instructive for us to redo this integral using letters instead of numbers. First, let us assume that n ≠ -1. The substitution is
u = ax + b
\frac{du}{dx} = a
dx = \frac{1}{a}du
which gives us
\int (ax + b)^n dx = \frac{1}{a}\int u^n dx
= \frac{1}{a}\Bigleft(\frac{u^{n+1}}{n+1}\Bigright) + C = \frac{(ax + b)^{n+1}}{a(n+1)} + C,
which looks just like the integral of x^n, but with x replaced by (ax + b), and the result divided by a.

We similarly get

\int (ax + b)^{-1} dx = \frac{1}{a}ln\|ax + b\| + C
\int e^{ax + b} dx = \frac{1}{a}e^{ax + b} + C.
Here is a summary of all of them:

Shortcuts: Integrals of Expressions Involving (ax + b)

If \int f(x) dx = F(x) + C, then

\int f(ax + b) dx = \frac{1}{a}F(ax + b) + C
for any constants a and b (a ≠ 0).
Regular Rule
\int x^n dx = \frac{x^{n+1}}{n+1} + C if (n ≠ -1)
Shortcut
\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C if (n ≠ -1)
Example
\int (3x - 1)^4 dx = \frac{(3x - 1)^5}{(3)(5)} + C = \frac{(3x - 1)^5}{15} + C
Technology: (3x-1)^3/9+C
Regular Rule
\int x^{-1} dx = ln\|x\| + C
Shortcut
\int (ax + b)^{-1} dx = \frac{1}{a}ln\|ax + b\| + C
Example
\int (4 - 3x)^{-1} dx = \frac{1}{-3}ln\|4 - 3x\| + C = -\frac{1}{3}ln\|4 - 3x\| + C
Technology: -(1/3)*ln(|4-3x|)+C   or   -(1/3)*ln(abs(4-3x))+C
Regular Rule
\int e^x dx = e^x + C
Shortcut
\int e^{ax + b} dx = \frac{1}{a} e^{ax + b} + C
Example
\int e^{-2x + 5} dx = \frac{1}{-2} e^{-2x + 5} + C = -\frac{1}{2} e^{-2x + 5} + C
Technology: -(1/2)*e^(-2x+5)+C
Regular Rule
\int \|x\| dx = \frac{1}{2}x\.\|x\| + C
Shortcut
\int \|ax + b\| dx = \frac{1}{2a}(ax + b)\.\|ax + b\| + C
Example
\int \|3 - x\| dx = \frac{1}{-2}(3 - x)\.\|3 - x\| + C = -\frac{1}{2}(3 - x)\.\|3 - x\| + C
Technology: -(3-x)*|3-x|/2+C   or   -(3-x)*abs(3-x)/2+C
Calculate each integral using the appropriate shortcut:

You can now try the exercises in Section 6.2 of , try some chapter review exercises, or go on to the next tutorial (press the link on the side).

Last Updated: April, 2010
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