# Solving Troublesome Integral: 100 \int \frac{h^2-20h}{(h-10)^2} dh

• Bashyboy
In summary: ClashPrefs%22+-%3E+%7B%22Math%22%7DIn summary, the conversation discusses a problem with computing a specific integral and presents a sample of work done to solve it. The Wolfram Alpha result is compared to the result obtained by the person and it is found that they differ by a constant. The conversation ends with a hint on how to approach the integral and a link to the correct result on Wolfram Alpha.
Bashyboy
Hello,

I have tried computing the following integral three times now, and I can not figure out what I am doing wrong.

The integral is $100 \int \frac{h^2-20h}{(h-10)^2} dh$, which wolfram calculates as $100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 (-h - \frac{100}{h-10} + c )$.

Here is one sample of my work:

Using integrating by parts, let $u = h^2 - 20h$, which becomes $\frac{du}{dh} = 2h - 20$. Multiplying both sides by the differential $u$, we get $\frac{du}{dh} dh = (2h-20)dh$. The definition of the differential of $u$ is $du = \frac{du}{dh} dh$. Making this substitution, $du = (2h - 20) dh$. Let $dv = \frac{dh}{(h-10)^2}$. Integrating both sides, $v = \frac{1}{h-10}$.

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[ (h^2-20h) \left(\frac{1}{h-10} \right) - \int \frac{2h-20}{h-10} dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int \frac{h-10}{h-10} dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2h + c \right]$

which is not the same...

Can anyone see what I might have done wrong?

Last edited:
Bashyboy said:
Hello,

I have tried computing the following integral three times now, and I can not figure out what I am doing wrong.

The integral is $100 \int \frac{h^2-20h}{(h-10)^2} dh$, which wolfram calculates as $100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 (-h - \frac{100}{h-10} + c$.

Here is one sample of my work:

Using integrating by parts, let $u = h^2 - 20h$, which becomes $\frac{du}{dh} = 2h - 20$. Multiplying both sides by the differential $u$, we get $\frac{du}{dh} dh = (2h-20)dh$. The definition of the differential of $u$ is $du \frac{du}{dh} dh$. Making this substitution, $du = (2h - 20) dh$. Let $dv = \frac{dh}{(h-10)^2}$. Integrating both sides, $v = \frac{1}{h-10}$.

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[ (h^2-20h) \left(\frac{1}{h-10} \right) - \int \frac{2h-20}{h-10} dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int \frac{h-10}{h-10} dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2 \int dh \right] \implies$

$100 \int \frac{h^2-20h}{(h-10)^2} dh = 100 \left[\frac{h^2-20h}{h-10} - 2h + c \right]$

which is not the same...

Can anyone see what I might have done wrong?

They differ by a constant. Divide ##\frac{h^2-20h}{h-10}## out using polynomial division. Using partial fractions to begin with would have led to an easier integration problem. BTW both solutions look to have a sign error.

Last edited:
Try using the substitution $$u = h-10$$
After some simple algebra, you should end up with the integral $$100∫\frac{u^2-100}{u^2} du$$ which should be fairly easy to evaluate, then you can substitute your h-10 back into your solution.

Last edited:
try using
$$100 \int \! \frac{h^2-20h}{(h-10)^2} \, \mathrm{d}h=100 \int \! \frac{(h-10)^2-100}{(h-10)^2} \, \mathrm{d}h$$

Bashyboy said:
Can anyone see what I might have done wrong?
Your particular mistake was integrating ##dv = \frac{dh}{(h-10)^2}## to get ##v = \frac{1}{h-10}##. That's pretty wrong. I don't know what Wolfram Alpha's problem is if you quoted it correctly.

Last edited:
I wouldn't consider that pretty wrong--perhaps just off the mark; after all, I am just missing a mere minus sign.

Bashyboy said:
I wouldn't consider that pretty wrong--perhaps just off the mark; after all, I am just missing a mere minus sign.

Why is the Wolfram Alpha answer you quoted also off by the same sign?

Why not simply calculating it carefully? It's not allowed to give the full solution in this forum, but here's a hint:

I'd use
$$\frac{h^2-20h}{(h-10)^2}=\frac{(h-10)^2-100}{(h-10)^2}=1-\frac{100}{(h-10)^2}.$$
This is trivial to integrate. Mathematica 9.0 under linux gets the correct result. So I wonder, why Wolfram alpha gets it wrong.

## 1. How do I simplify this integral?

In order to simplify this integral, you can use the method of partial fractions. First, factor out the numerator and denominator of the fraction. Then, set up the partial fraction equation and solve for the constants. Finally, integrate each term separately to get the final answer.

## 2. Can this integral be solved using substitution?

Yes, this integral can also be solved using substitution. Let u = h-10 and du = dh. Then, the integral becomes 100 \int \frac{(u+10)^2-20(u+10)}{u^2} du. Simplify the expression and then integrate to find the final answer.

## 3. Is there a specific technique to use for this type of integral?

Yes, this integral falls under the category of rational functions and can be solved using the method of partial fractions or substitution. Other techniques that can be used for rational functions include integration by parts and trigonometric substitution.

## 4. What is the final answer to this integral?

The final answer to this integral is 100 \int \frac{h^2-20h}{(h-10)^2} dh = -100h + C + 100ln|h-10| + 1000. The constant C represents the constant of integration and can be determined by using initial conditions or boundaries.

## 5. Can this integral be solved using software or calculators?

Yes, this integral can be solved using software or calculators that have integral-solving capabilities. However, it is important to note that these tools may not always provide the most simplified answer and it is always beneficial to understand the steps and techniques used to solve the integral manually.

Replies
10
Views
1K
Replies
15
Views
6K
Replies
7
Views
2K
Replies
2
Views
1K
Replies
4
Views
1K
Replies
8
Views
1K
Replies
3
Views
2K
Replies
22
Views
2K
Replies
12
Views
2K
Replies
20
Views
978