Good Question 13

Let’s end the year with this problem that I came across a while ago in a review book:

Integrate \int{x\sqrt{x+1}dx}

It was a multiple-choice question and had four choices for the answer. The author intended it to be done with a u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.

    1. Find the antiderivative using a u-substitution.
    2. Find the antiderivative using integration by parts.
    3. Find the antiderivative using a different u-substitution.
    4. Find the antiderivative by adding zero in a convenient form.

Your answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by

  1. “Simplifying” your answer to 2 and get a third form of the answer.
  2. “Simplifying” your answer to 1, 3, 4 and get that third form again.

Give it a try before reading on. The solutions are below the picture.

Method 1: u-substitution

Integrate \int{x\sqrt{x+1}dx}


\int{x\sqrt{x+1}dx=}\int{\left( u-1 \right)\sqrt{u}}du=\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\tfrac{5}{2}{{u}^{5/2}}-\tfrac{3}{2}{{u}^{3/2}}

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

Method 2: By Parts

Integrate \int{x\sqrt{x+1}dx}


dv=\sqrt{x+1}dx,v=\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}

\int{x\sqrt{x+1}dx}=\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\int{\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}dx}=

\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C

Method 3: A different u-substitution

Integrate \int{x\sqrt{x+1}dx}


du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu

2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

This gives the same answer as Method 1.

Method 4: Add zero in a convenient form.

Integrate \int{x\sqrt{x+1}dx}

\int{x\sqrt{x+1}}dx=\int{x\sqrt{x+1}+\sqrt{x+1}-\sqrt{x+1} dx=}

\int{\left( x+1 \right)\sqrt{x+1}-\sqrt{x+1}}dx=

\int{{{\left( x+1 \right)}^{3/2}}-{{\left( x+1 \right)}^{1/2}}dx}=

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

This, also, gives the same answer as Methods 1 and 3.

So, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?

No, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of {{\left( x+1 \right)}^{3/2}}. (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)

Simplify the answer for Method 2:

\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C=

\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( x-\tfrac{2}{5}\left( x+1 \right) \right)+C=

\displaystyle \tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( \frac{5x-2x-2}{5} \right)+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C

Simplify the answer for Methods 1, 3, and 4:

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3\left( x+1 \right)-5 \right)+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C

So, same answer and same constant.

Is this a good question? No and yes.

As a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.

From another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students’ perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (… maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.

My next post will be after the holidays.

Happy Holidays to Everyone!




Since a lot of classes start integration with antidifferentiation, I’ll discuss that first. If you decide to go with one of the other plans mentioned in my last post, then file this away for later.

The key to finding antiderivatives is pattern recognition.

The simplest integrals are those that follow directly from derivatives such as. Students just need to recognize that the cosine is the derivative of the sine so \int{\cos \left( x \right)dx}=\sin \left( x \right)+C. We wish they were all that simple.

Something like \int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx} is more complicated. Is the integrand the derivative of \tfrac{1}{2}{{\tan }^{2}}\left( x \right)  or of \tfrac{1}{2}{{\sec }^{2}}\left( x \right)? The answer is yes.

The method called u-substitution helps in identifying the pattern of the integrand.  To use this method identify some part of the integrand as a function which you call u and then calculate du and hope that the du is the rest of the integrand.

For \int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx} you can try u = tan(x) and the du = sec2(x)dx and after making the substitutions:

\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}=\int{udu}=\tfrac{1}{2}{{u}^{2}}=\tfrac{1}{2}{{\tan }^{2}}\left( x \right)+C.

On the other hand we could try u = sec(x) so that du = tan(x)sec(x)dx and

\int{\tan \left( x \right){{\sec }^{2}}\left( x \right)dx}=\int{\sec (x)\left( \tan (x)\sec (x)dx \right)}

=\int{udu}=\tfrac{1}{2}{{u}^{2}}=\tfrac{1}{2}{{\sec }^{2}}\left( x \right)+C*

(The + C* here not the same as the + C in the expression above: C* = C – ½.)

Either way students need to recognize that part of the integrand is the chain rule contribution to the derivative of the other part of the integrand. The usual three steps to acquire these pattern recognition skills are practice, practice, practice.

Another detail of u-substitution is handling missing constants. For example, \int{\cos (2x)dx} .

Make the substitution u = 2x and then calculate du = 2dx so dx=\tfrac{1}{2}du. Then making these substitutions

\int{\cos (2x)dx}=\int{\cos \left( u \right)\left( \tfrac{1}{2}du \right)=}\tfrac{1}{2}\int{\cos \left( u \right)du=\tfrac{1}{2}\sin \left( u \right)}.

Then back-substituting: \int{\cos (2x)dx}=\tfrac{1}{2}\sin \left( 2x \right)+C.

For simple u-substitutions like this I find it easier to think, and not write, u = 2x so du = 2dx and then write this multiplying by ½ outside the integral sign to account for the extra factor of 2.

\int{\cos \left( 2x \right)dx=\tfrac{1}{2}\int{\cos \left( 2x \right)\left( 2dx \right)}}=\tfrac{1}{2}\sin \left( 2x \right)+C.

This idea only works with missing constants, since only constants can be moved in and out of integrands.

I explain the need for the constant to students by saying that the 2 appears (as if by magic) from the Chain Rule when you differentiate and therefore it must be present to disappear back into the Chain Rule when you antidifferentiate.

Point out to students that integration is very different than differentiation. Differentiation is pretty straightforward; you know what to do when you need to find a derivative. Antidifferentiation is more complicated since recognizing the form or pattern is necessary. The simpler looking integral \int{\ln \left( x \right)}dx is really more difficult than \int{\tfrac{\ln \left( x \right)}{x}dx}.

Finally, if you are teaching antiderivatives before beginning integration, when you get to definite integrals, you will have to remember to show students how to handle the limits of integration by using the same substitution.

For Advanced Placement AB calculus courses these (integrals that follow directly from derivatives and u-substitutions) are the only methods of integration tested. BC students also need to know integration by parts and partial fraction decomposition. I will discuss these later.