*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I've been trying to integrate $\int \frac{x^2}{x-1} \dx$ for what feels like days and I'm completely stuck. I worry that I'm not cut out for A-level!

-- Calculus: Amazingly Useful, Can Hurt You

Hi, there, CAUCHY -- and don't worry, not being able to do something doesn't mean you're not cut out for A-level. In fact, trying hard and then asking for help is exactly the kind of skill I try to foster in my students -- so well done for having a stab.

With a question like this, there are (at least) two options: an easy one (the $u$-substitution) and a more involved one (the division method).

### The $u$-substitution

The thing to ask yourself when contemplating a substitution is, 'what's the ugly thing here?'. Clearly enough, it's the $x-1$ on the bottom of the fraction, so let's say $u = x-1$ and -- to howls of anguish from the purists, $\d u = \d x$.

Then our integral becomes $\int \frac{x^2}{u} \d u$. But wait! There's still a nasty $x$ knocking around -- fortunately, we can rearrange our substitution to say $x = u+1$, so the top can be written as $u^2 + 2u + 1$.

$\int \frac{u^2 + 2u + 1}{u} \d u = \int u + 2 + \frac 1u \d u$, which is much nicer. You get $\frac {u^2}{2} + 2u + \ln\left| u\right| + C$, nice as you like.

One last thing, though: we introduced the $u$ ourselves, so we need to turn it back into an $x$, giving

$\frac{(x-1)^2}{2} + 2(x-1) + \ln \left| x-1 \right| + C$.

(A smart-arse would tidy that up as $\frac {x^2}2 + x + \ln \left| x-1 \right| + c$. Can you see why that works?)

### The division method

An alternative, which may be more comfortable, even if it's longer-winded, is to simplify the fraction by division: $\frac{x^2}{x-1} = x+1 + \frac 1{x-1}$, which is almost trivial to integrate, giving $\frac {x^2} 2 + x + \ln \left| x-1 \right| + C$ -- the same answer as the other way!

Hope that helps you find some rest!

-- Uncle Colin

* Edited 2015-07-22 to change a minus sign. Thanks, @realityminus3!

* Edited 2016-05-18 to change brackets to absolute values. Thanks to Reuben for reminding me of the post!

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## MathbloggingAll

Ask Uncle Colin: A Sticky Integral http://t.co/5nO6aYlu8z