U-Substitution

THE METHOD OF U-SUBSTITUTION


The following problems involve the method of u-substitution. It is a method for finding antiderivatives. We will assume knowledge of the following well-known, basic indefinite integral formulas : The method of u-substitution is a method for algebraically simplifying the form of a function so that its antiderivative can be easily recognized. This method is intimately related to the chain rule for differentiation. For example, since the derivative of ex is

$ D \{ e^x \} = e^x $ ,

it follows easily that

$ \displaystyle{ \int e^x \,dx } = e^x + C $ .

However, it may not be obvious to some how to integrate

$ \displaystyle{ \int (2x+2) e^{ x^2 + 2x + 3 } \,dx } $ .

Note that the derivative of $ \displaystyle{ e^{ x^2 + 2x + 3 } } $ can be computed using the chain rule and is

$ \displaystyle{ D \{ e^{ x^2 + 2x + 3 } \} = e^{ x^2+2x+3 } \ D \{ x^2+2x+3 \} } $

$ = \displaystyle{ e^{ x^2+2x+3 } (2x+2) } $

$ = \displaystyle{ (2x+2) e^{ x^2+2x+3 } } $ .

Thus, it follows easily that

$ \displaystyle{ \int (2x+2) e^{ x^2+2x+3 } \,dx } = e^{ x^2+2x+3 } + C $ .

This is an illustration of the chain rule "backwards". Now the method of u-substitution will be illustrated on this same example. Begin with

$ \displaystyle{ \int (2x+2) e^{ x^2 + 2x + 3 } \,dx } $ ,

and let

u = x2+2x+3 .

Then the derivative of u is

$ \displaystyle{ { du \over dx } } = 2x+2 $ .

Now "pretend" that the differentiation notation $ \displaystyle{ { du \over dx } } $ is an arithmetic fraction, and multiply both sides of the previous equation by dx getting

$ \displaystyle{ { du \over dx } \ dx } = (2x+2) dx $

or

du = (2x+2) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

$ \displaystyle{ \int (2x+2) e^{ x^2+2x+3 } \,dx }
= \displaystyle{ \int e^{ x^2+2x+3 } (2x+2) \,dx }$

$ = \displaystyle{ \int e^{ u } \,du }$

= e u + C

= e x2+2x+3 + C .

Of course, it is the same answer that we got before, using the chain rule "backwards". In essence, the method of u-substitution is a way to recognize the antiderivative of a chain rule derivative. Here is another illustraion of u-substitution. Consider

$ \displaystyle{ \int { x^2+1 \over x^3+3x } \,dx } $ .

Let

u = x3+3x .

Then (Go directly to the du part.)

du = (3x2+3) dx = 3(x2+1) dx ,

so that

(1/3) du = (x2+1) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

$ \displaystyle{ \int { x^2+1 \over x^3+3x } \,dx }
= \displaystyle{ \int { 1 \over x^3+3x } \ (x^2+1) \,dx }$

$ = \displaystyle{ \int { 1 \over u } (1/3) \,du } $

$ = \displaystyle{ (1/3) \int { 1 \over u } \,du } $

$ = (1/3) \ln\vert u\vert + C $

$ = (1/3) \ln\vert x^3+3x\vert + C $ .

Most of the following problems are average. A few are challenging. Make careful and precise use of the differential notation dx and du and be careful when arithmetically and algebraically simplifying expressions.



The following problems require u-substitution with a variation. I call this variation a "back substitution". For example, if u = x+1 , then x=u-1 is what I refer to as a "back substitution".


Click HERE to return to the original list of various types of calculus problems.


Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu


Duane Kouba
1999-05-04