# Integration tables

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• Mar 4th 2008, 06:07 PM
ebonyscythe
Integration tables
I was given the following problem:

A. Use Derive to evaluate the integral $\displaystyle \int x^5*(x^2+1)^{1/2} dx$

B. Use integration tables in the back of your book to evaluate the integral from part A. State the table entry used and show all work. After you find this antiderivative by hand, show that it is equal to the antiderivative given by derive.

While using derive, I obtained this as the answer for part A:

= $\displaystyle \frac{(x^2+1)^{1/2}(15x^6+3x^4-4x^2+8)}{105}$

But using my Ti-89 I got this instead:

= $\displaystyle \frac{(x^2+1)^{3/2}(15x^4-12x^2+8)}{105}$

So I'm not sure what's going on there, but I went ahead and tried out an integration table that looked useful:

$\displaystyle \int u^n(bu+a)^{1/2} du = \frac{2}{b(2n+3)}(u^n(a+bu)^{3/2} - na \int u^{n-1}(bu+a)^{1/2} du$

However, when I use u=x, n=5 and b=x the integral at the end looks just as messy as what I started with. Do I really have to use the integration table again on the last integral part of the integration table? Am I doing it wrong?

Edit: I found an online pdf of the integration tables I have access to at this site: http://teachers.sduhsd.k12.ca.us/abr...lesStewart.pdf
• Mar 4th 2008, 06:25 PM
topsquark
Quote:

Originally Posted by ebonyscythe
= $\displaystyle \frac{(x^2+1)^{1/2}(15x^6+3x^4-4x^2+8)}{105}$

= $\displaystyle \frac{(x^2+1)^{3/2}(15x^4-12x^2+8)}{105}$

Note that $\displaystyle (x^2 + 1)^{3/2} = (x^2 + 1)(x^2 + 1)^{1/2}$. So what is
$\displaystyle (x^2 + 1)(15x^4 - 12x^2 + 8)$

-Dan
• Mar 4th 2008, 06:49 PM
ebonyscythe
Ah, so they are equal. I didn't look at it that way at all. Thanks for that, now I can put that part aside.

Can anyone help with the integration tables next?
• Mar 4th 2008, 07:17 PM
topsquark
Quote:

Originally Posted by ebonyscythe
So I'm not sure what's going on there, but I went ahead and tried out an integration table that looked useful:

$\displaystyle \int u^n(bu+a)^{1/2} du = \frac{2}{b(2n+3)}(u^n(a+bu)^{3/2} - na \int u^{n-1}(bu+a)^{1/2} du$

However, when I use u=x, n=5 and b=x the integral at the end looks just as messy as what I started with. Do I really have to use the integration table again on the last integral part of the integration table? Am I doing it wrong?

Edit: I found an online pdf of the integration tables I have access to at this site: http://teachers.sduhsd.k12.ca.us/abr...lesStewart.pdf

$\displaystyle \int x^5*(x^2+1)^{1/2} dx$

Let $\displaystyle u = x^2$. Then your integral becomes
$\displaystyle = \int u^{5/2}(u + 1)^{1/2}~\frac{1}{2 u^{1/2}}~du$

$\displaystyle = \frac{1}{2} \int u^2(u + 1)^{1/2}~du$

So n = 2, b = 1, and a = 1.

Thus according to the table:
$\displaystyle \int u^2(u + 1)^{1/2}~du = \frac{2}{7} \left ( u^2(1+u)^{3/2} - 2 \int u(u+1)^{1/2}~du \right )$

and
$\displaystyle \int u(u+1)^{1/2}~du = \frac{2}{5} \left ( u(1+u)^{3/2} - \int (u+1)^{1/2} ~du \right )$

and
$\displaystyle \int (u+1)^{1/2} ~du = \frac{2}{3}(u + 1)^{3/2}$

So
$\displaystyle \int u^2(u + 1)^{1/2}~du = \frac{22}{105}u^2(1+u)^{3/2}$
after a lot of simplifying.

Thus
$\displaystyle \int x^5(x^2+1)^{1/2}~dx = \frac{11}{105}x^4(x^2 + 1)^{3/2}$

Which is obviously not your answer. Unless I missed some factors something is wrong in the table.

-Dan
• Mar 4th 2008, 07:22 PM
ebonyscythe
I see what I did wrong. I didn't perform the U-substitution right away, like you did, and must've made it way harder on myself. Thank you so much for the help, I think that's exactly what I need... now to absorb it.
• Mar 4th 2008, 07:53 PM
ebonyscythe
Quote:

Originally Posted by topsquark
$\displaystyle = \frac{1}{2} \int u^2(u + 1)^{1/2}~du$

So n = 2, b = 1, and a = 1.

Thus according to the table:
$\displaystyle \int u^2(u + 1)^{1/2}~du = \frac{2}{7} \left ( u^2(1+u)^{3/2} - 2 \int u(u+1)^{1/2}~du \right )$

Question, what happened to the 1/2 that was in front of the first integral?
• Mar 5th 2008, 09:31 AM
topsquark
Quote:

Originally Posted by ebonyscythe
Question, what happened to the 1/2 that was in front of the first integral?

The factor of 1/2 relates the x integral to the u integral.

-Dan