Tricky indefinite integral problem

In summary: Your table should also have a formula for ##\int du/(u^2-a^2)##, which you can use to integrate ##1/(u^2-2^2)##.Homework Statement Ok the problem is:∫-1/(4x-x^2) dxThe answer in the back of the book is:(1/4)ln(abs((x-4)/x))) + CHomework EquationsI think this would be used somehow:∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + CI assume you got that from a table you are allowed to use?
  • #1
zak1989
6
0

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 :)
 
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  • #2
Use partial fractions:
[tex]
\frac{-1}{4x-x^2} = \frac{1}{x(x-4)}
[/tex]
Now find A and B where
[tex]
\frac{1}{x(4-x)} \equiv \frac Ax + \frac B{x-4}
[/tex]

You should now be able to integrate each term separately.

(You can write
[tex]
\frac{1}{x^2 - 4x} = \frac{1}{(x - 2)^2 - 2^2}
[/tex]
and substitute [itex]u = x - 2[/itex], but the partial fraction approach is the usual method.)
 
Last edited:
  • #3
zak1989 said:

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.
zak1989 said:
∫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 :)
 
  • #4
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.
 
  • #5
A trig substitution should work. Have you learned that technique yet?
 
  • #6
zak1989 said:

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?

zak1989 said:
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.
 

1. What is an indefinite integral?

An indefinite integral is a mathematical operation used to find the most general antiderivative of a function. It is used to reverse the process of differentiation, where the derivative of a function is found.

2. How do I solve a tricky indefinite integral problem?

The best approach to solving a tricky indefinite integral problem is to first identify the type of function and then use integration techniques such as substitution, integration by parts, or trigonometric substitution. It is also important to remember to check your solution by differentiating it to ensure it matches the original function.

3. Can I use a calculator to solve an indefinite integral?

Yes, there are many online or handheld calculators that can solve indefinite integrals. However, it is important to understand the steps involved in solving the integral, as calculators may not always provide the most accurate or simplified solution.

4. What are common mistakes to avoid when solving indefinite integrals?

Some common mistakes to avoid when solving indefinite integrals include missing a constant of integration, forgetting to apply integration rules, and making errors in algebraic simplification. It is also important to check the limits of integration and ensure they are correct.

5. Are there any tips for solving tricky indefinite integrals?

One helpful tip for solving tricky indefinite integrals is to practice and familiarize yourself with different integration techniques. It is also beneficial to break down complex integrals into smaller, more manageable parts. Additionally, checking your solution using differentiation can help catch any mistakes.

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