Tricky indefinite integral problem

[zak1989]

Homework Statement

Ok the problem is:
∫-1/(4x-x^2) dx

The answer in the back of the book is:
(1/4)ln(abs((x-4)/x))) + C

Homework Equations

I think this would be used somehow:
∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C

The Attempt at a Solution

∫-1/(4x-x^2) dx

∫-1/(-x^2+4x) dx

∫-1/-(x^2-4x) dx

∫1/(x^2-4x) dx

∫1/(x^2-4x+4-4) dx

∫1/((x-2)^2-4) dx

∫1/((x-2)^2-2^2) dx

At this point I'm stumped. No matter what I try I can seem to get it to look like the book answer.

If I move the bottom part around and use u-substitution I get:

∫1/(u^2-2^2) du let u=x-2 and du = dx

∫1/(-2^2+u^2) du

∫1/-(2^2-u^2) du

∫-1/(2^2-u^2) du

-∫1/(2^2-u^2) du

I will get a negative number which isn't what I want. What am I doing wrong? I would greatly appreciate any help. Thanks :)

 

[pasmith]

Use partial fractions:
<br /> \frac{-1}{4x-x^2} = \frac{1}{x(x-4)}<br />
Now find A and B where
<br /> \frac{1}{x(4-x)} \equiv \frac Ax + \frac B{x-4}<br />

You should now be able to integrate each term separately.

(You can write
<br /> \frac{1}{x^2 - 4x} = \frac{1}{(x - 2)^2 - 2^2}<br />
and substitute u = x - 2, but the partial fraction approach is the usual method.)

 

[Mark44]

Homework Statement

Ok the problem is:
∫-1/(4x-x^2) dx

The answer in the back of the book is:
(1/4)ln(abs((x-4)/x))) + C

Homework Equations

I think this would be used somehow:
∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C

The Attempt at a Solution

∫-1/(4x-x^2) dx

∫-1/(-x^2+4x) dx

∫-1/-(x^2-4x) dx

∫1/(x^2-4x) dx

∫1/(x^2-4x+4-4) dx

∫1/((x-2)^2-4) dx

∫1/((x-2)^2-2^2) dx

At this point I'm stumped. No matter what I try I can seem to get it to look like the book answer.

If I move the bottom part around and use u-substitution I get:

∫1/(u^2-2^2) du let u=x-2 and du = dx

Use partial fractions to decompose the integrand into A/(u - 2) + B/(u + 2).

IOW, solve this equation for A and B:
$$\frac{1}{(u - 2)(u + 2)} = \frac{A}{u - 2} +\frac{B}{u + 2} $$

The equation above actually must be identically true (true for any values of u other than 2 and -2.

∫1/(-2^2+u^2) du

∫1/-(2^2-u^2) du

∫-1/(2^2-u^2) du

-∫1/(2^2-u^2) du

I will get a negative number which isn't what I want. What am I doing wrong? I would greatly appreciate any help. Thanks :)

 

[zak1989]

Thanks for the help guys, but how would I go about doing it without partial fractions? I'm not supposed to use that technique yet. Sorry I forgot to mention that in my post.

 

[Mark44]

A trig substitution should work. Have you learned that technique yet?

 

[LCKurtz]

Homework Statement

Ok the problem is:
∫-1/(4x-x^2) dx

The answer in the back of the book is:
(1/4)ln(abs((x-4)/x))) + C

Homework Equations

I think this would be used somehow:
∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C

I assume you got that from a table you are allowed to use?

Thanks for the help guys, but how would I go about doing it without partial fractions?

Try pasmith's suggestion $u=x-2$ and $a=2$ in your table formula above.