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I am trying to find the integral of:

$$\int \frac{dx}{x\sqrt{x-1}}$$

if I do this $u$-sub:

$u = \sqrt{x-1}$ and $u^2 = x-1$ and $x = u^2 + 1$ and $\frac{dx}{du} = 2u$ and $dx = 2udu$

Then I get:

$$\int \frac{2udu}{(u^2+1)u}$$

if I do a partial fraction of this I wind up stuck and I wonder why:

$$\frac{2udu}{(u^2+1)u} = \frac{A}{u} + \frac{Bu + C}{u^2+1}$$

$$2u = Au^2 + A + Bu^2 + Cu$$

$$2u = u^2(A+B) + Cu + A$$

$$C = 2$$

$$A=0$$ $$B=0$$

But that doesn't seem right because then I'd get $\frac{2}{u^2+1}$ which is not equal to the original fraction.

But if I do some cancelling:

$$\frac{2udu}{(u^2+1)u} = \frac{2}{u^2+1} = 2 \frac{1}{u^2 + 1}$$

So then:

$$\int \frac{2udu}{(u^2+1)u}$$

$$2 \int \frac{1}{u^2 + 1} du$$

$$ = 2\arctan(u)$$

$$2\arctan(\sqrt{x-1}) + C$$

Does not simplifying ruin partial fractions?

EDIT

ohhhhhh I get it. It's the same thing! Reduced or using partial fractions, I wind up with $\int \frac{2}{u^2+1}$. Is that right?$

Jwan622
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    ohhhhhh I get it. It's the same thing! Reduced or using partial fractions, I wind up with $\frac{2}{u^2+1}. Is that right?$ – Jwan622 Feb 20 '19 at 03:06
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    You don't need partial fractions. You can divide numerator and denominator by $u$ to get the same place. Partial fractions worked fine, though. Yes, the integrand is $\frac 2{u^2+1}$ – Ross Millikan Feb 20 '19 at 03:09

2 Answers2

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As you see in your workings, you didn't cancel $u$ from your fraction so the $u$ part of the partial fractions had a coefficient of $0$, hence it doesn't exist in the simplification, which is what's you'd expect considering it can be cancelled.

Rhys Hughes
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I think the answer has already clicked for you, but just to clarify: yes, both your work using partial fractions and your work cancelling $u$ in the numerator and denominator of the integrand $\frac {2u}{(u^2+1)u}$ lead to the same correct simplified expression of $\frac 2{u^2+1}$ for the integrand, from which you correctly finish evaluating the integral.

Often it can be quite easy to slightly confuse oneself when manipulating algebraic expressions, especially when performing integration. Here, despite the fact that you struggled reaching a solution, all of the 'real' mathematics you did was (essentially) correct! You were just stopped at the end of your partial fraction work by a slight mental blip.

I find it's easiest to avoid this if I set out my work as clearly and simply as possible. The more parts there are to the algebraic expressions I'm working with and the more effort I have to use to read what I've written the more likely I am to make a mistake or momentarily not think properly about what I'm doing. I mention that your work was only essentially correct; you haven't been completely disciplined with your handling of $du$. Namely, $$\frac {2udu}{(u^2+1)u} \neq \frac Au + \frac {Bu+C}{u^2 +1}$$ and $$\frac {2udu}{(u^2+1)u} \neq \frac 2{u^2+1}$$ unlike what you've written. In both of these cases $du$ has disappeared from the right hand side of the equation!

To make things clearer and simpler, once you've reached the expression $\int \frac {2udu}{(u^2+1)u}$ note that you're just trying to manipulate the integrand $\frac {2u}{(u^2+1)u}$ into a form that you can integrate and so it's easiest to leave $du$ entirely out that manipulation work. If you start by instead trying to manipulate $$\frac {2u}{(u^2+1)u}$$ you'll probably realise straight away that you can cancel $u$ from the numerator and denominator and so avoid starting going down the much lengthier (but just as valid) route of using partial fractions.

(As a final note, be careful at which point you add the arbitrary constant when integrating by substitution. You should have added it a line earlier than you did. Your working should read $$2\int \frac 1{u^2+1}du = 2\arctan (u) + C$$ This is because the arbitrary constant is a result of performing the integration, not of substituting back $\sqrt {x-1}$ for $u$, as is implied by your working.)

user646558
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