# Troublesome Integral

1. Mar 16, 2014

### 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: Mar 17, 2014
2. Mar 16, 2014

### Dick

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: Mar 16, 2014
3. Mar 16, 2014

### vanceEE

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: Mar 16, 2014
4. Mar 16, 2014

### lurflurf

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

5. Mar 16, 2014

### Dick

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: Mar 16, 2014
6. Mar 17, 2014

### Bashyboy

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

7. Mar 17, 2014

### Dick

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

8. Mar 17, 2014

### vanhees71

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.

9. Mar 17, 2014