You'l just love this intergral...?

Question

How would you intergrate:

1/((a-x)(2a-x))

I can never remember how to do it wen its 1 over something!

is it ln(a-x)(2a-x) ...?

Answer 1

You have to separate the denominator by using partial fractions. Make the expression equal to s/(a-x) +t/(2a-x), find the corresponding values for s and t and integrate. You will then end up with ln something or other - try it and see.

Answer 2

You have to use partial fractions. Because they're linear factors (the power of x is one), this one fraction can be split up into two fractions as follows:

1/[(a-x)(2a-x)] = A/(a-x) + B/(2a-x), for some constants A, B.

If we integrate the newly made fractions and solve for the constants later, we get

Integral (A/(a-x) + B/(2a-x))dx

We can split this up into two integrals.

Integral (A/(a-x))dx + Integral (B/(2a-x))dx

And we pull out constants out of the integrals

A * Integral (1/(a-x))dx + B * Integral (1/(2a-x))dx

To integrate when the bottom is a linear factor, it will most definitely involve ln, since the integral of 1/x is ln|x|+C. We do a similar thing here, except we have to keep in mind the value to offset the result of the chain rule.

For instance, if we guess ln|a-x| as the integral of the first fraction, taking the derivative would yield 1/(a-x) * (-1), so we have to add this value to the integral to offset the chain rule (a bit confusing, I know, but gets easier to understand with practice). Ultimately you'll obtain this result for the integral:

A * (-ln |a - x|) + B * (-ln |2a - x|) + C
-A (ln |a-x|) - B (ln |2a-x|) + C

That's our answer. But it's not finished because we haven't even begun to solve for A and B. The process to solve for A and B is below.

Recall that

1/[(a-x)(2a-x)] = A/(a-x) + B/(2a-x)

What we want to do at this point is multiply the equation by (a-x)(2a-x). This means one of those will cancel out the other on the right hand side, and we would be left with

1 = A(2a - x) + B(a - x)

Remember that the above equation is true FOR ALL X. This means we can substitute ANYTHING for x. For cancellation purposes and for getting the values for A and B, let x = 2a and then afterward, let x = a.

If x = 2a, then we get the equation
1 = A(2a - 2a) + B(a - 2a)
1 = 0 + B(-a)
B = -1/a

If x = a, then we get the equation
1 = A(2a - a) + B(a - a)
1 = A(a) + 0
1 = Aa, therefore
A = 1/a

Now that we have our values, let's refer back to our answer.

-A (ln |a-x|) - B (ln |2a-x|) + C

Substitue A = 1/a and B = (-1/a), to get

(1/a) ln|a-x| - (-1/a)(ln|2a-x|) + C

or

(1/a) ln|a-x| + (1/a)(ln|2a-x|) + C

If we really wanted to get technical and reduce it further (It's possible!), we can factor 1/a out of both and then use log properties to combine it into a single log.

(1/a) [ln|a-x| + ln|2a-x|] + C
(1/a) [ln[(a-x)(2a-x)] + C

Answer 3

very late.. same idea as 'puggy'
partial fractions> 1/(a-x)(2a-x) identical to A/(a-x)+B/(2a-x)
clearing fractions gives A(2a-x)+B(a-x)=1
identity true for all x, subst x=2a and x=a successively yields B=-1/a and A=1/a
So integral becomes
1/a(int(1/(a-x)-int(1/(2a-x))
> 1/a(-ln(a-x)+ln(2a-x))
> ln{(2a-x)/(a-x)}^1/a