# Evaluating an integral involving a fraction

1. Jul 17, 2011

### sara_87

1. The problem statement, all variables and given/known data

I want to integrate:

$\int (\frac{x-a}{b-x})^{c/d}dx$

2. Relevant equations

given:
c<d

3. The attempt at a solution

I have no idea which integration method to use. i tried to make the integrand look easier by rewriting it as:

$(1+\frac{b-a}{x-b})^{c/d}$

but this still doesnt help me.
any ideas will be very much appreciated. thank you.

2. Jul 17, 2011

### hunt_mat

Have you tried the obvious one?
$$u=\frac{x-a}{b-x}$$

3. Jul 17, 2011

### sara_87

after using the substitution:

$u=\frac{x-a}{b-x}$,

we find the derivative of u with respect to x so that:

$du=\frac{b-a}{(b-x)^2}dx$
(using the quotient rule for differentiation),

then, substituting this into the original integral, we have:

$(b-a)\int\frac{u^{c/d+1}}{(u+1)^2}du$

and from here, i can't seem to think of another substitution.
any ideas?

4. Jul 17, 2011

### 1MileCrash

Is c supposed to represent the dummy constant when evaluating an indefinite integral, or is it just a generic variable here like a and b?

5. Jul 17, 2011

### sara_87

it is a generic variable like a and b.

6. Jul 17, 2011

### hunt_mat

Hmm, that ends up quite messy and not very helpful in the slightest (but you did it correctly), you did make another (very nice) observation that you integrand becomes:
$$\left(1+\frac{b-a}{b-x}\right)^{\frac{c}{d}}$$
Now I would use the following substitution:
$$u=\frac{1}{b-x}$$

Last edited: Jul 17, 2011
7. Jul 17, 2011

### sara_87

I tried this substitution u=1/(x-b),
so du=ln(x-b)dx

so, substituting this into the integral, we have:

$\int\frac{(1+(b-a)u)^{c/d}}{-ln(u)}du$

and... also, i am stuck on what to chose for the next substitution.

8. Jul 17, 2011

### hunt_mat

That is not what you get, you integrated instead of differentiated, what is du?

9. Jul 17, 2011

### sara_87

my mistake.

du=-1/(x-b)^2 dx

so it should be:

$\int\frac{(1+(b-a)u)^{c/d}}{u^2}du$

10. Jul 17, 2011

### hunt_mat

Any thoughts now?

11. Jul 18, 2011

### sara_87

i can't think of a further useful substitution, because we have the power of c/d.