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

[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?

 

[Dick]

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.

 

[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.

 

[lurflurf]

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

 

[Dick]

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.

 

[Bashyboy]

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

 

[Dick]

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?

 

[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.

 

[Bashyboy]

The reason for that is, that I was a ninny and typed in the wolfram's results incorrectly. Here is a link to wolfram alpha, with the correct result, hopefully: https://www.wolframalpha.com/input/?i=int+(h^2-20h)/(h-10)^2