Integration Help 2: Solving Int. w/ Unknown Method

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In summary, we discussed the integral $\int\frac{2x^2}{2x^2-1}dx$ and different methods that could be used to solve it. These include using a substitution and the method of partial fraction decomposition. The latter method involves rewriting the fraction as a sum of simpler fractions and solving for the unknown constants using various techniques such as equating coefficients or evaluating the fraction at specific values of x.
  • #1
NotaMathPerson
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I need some help with this integran


$$\int\frac{2x^2}{2x^2-1}dx$$I can't seem to solve this using the techniques that I know.
What method should I use?

 
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  • #2
Observe that:

\(\displaystyle \frac{2x^2}{2x^2-1}=\frac{2x^2-1+1}{2x^2-1}=1+\frac{1}{2x^2-1}\)

Can you now proceed with a substitution?
 
  • #3
There are several different ways to do this but the first thing I would do is multiply both numerator and denominator by -1: [tex]\frac{-2x^2}{1- 2x^2}dx= -\frac{2x^2}{1- 2x^2}dx[/tex]. I would do that because the denominator, [tex]1- 2x^2[/tex], reminds me of a trig substitution. [tex]sin^2(\theta)+ cos^2(\theta)= 1[/tex] so [tex]cos^2(\theta)= 1- sin^2(\theta)[/tex]. If we let [tex]\sqrt{2}x= sin(\theta)[/tex] then [tex]\sqrt{2}dx= cos(\theta)d\theta[/tex] so [tex]dx= \frac{1}{\sqrt{2}}cos(\theta)d\theta[/tex], and [tex]2x^2= sin^2(\theta).

We have [tex]\int \frac{2x^2}{1- 2x^2}dx= -\int \frac{2x}{2x^2- 1}dx[/tex][tex]= -\int\frac{sin^2(\theta)}{cos^2(\theta)}\left(\frac{1}{\sqrt{2}}cos(\theta)d\theta\right)[/tex][tex]= -\frac{1}{\sqrt{2}}\int\frac{sin^2(\theta)}{1- sin^2(\theta)}cos(\theta) d\theta[/tex] and now the substitution [tex]u= sin(\theta)[/tex] changes this to a "rational" function: [tex]-\frac{1}{\sqrt{2}}\int \frac{u^2}{1- u^2}du[/tex].
 
  • #4
MarkFL said:
Observe that:

\(\displaystyle \frac{2x^2}{2x^2-1}=\frac{2x^2-1+1}{2x^2-1}=1+\frac{1}{2x^2-1}\)

Can you now proceed with a substitution?

Do you mean $$u=2x^2-1$$

$$du=4xdx$$ there's still an x.
 
  • #5
I was thinking more along the lines of a trig. or hyperbolic trig. substitution. You could also try partial fraction decomposition. :)
 
  • #6


Please check my solution.

$$-\frac{1}{\sqrt{2}}\int\frac{u^2}{1-u^2}\,du$$Using partial fraction decomposition

$$\frac{u^2}{1-u^2}=\frac{u^2}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$$

u=1; A = 0.5
u=-1; B =0.5

$$\frac{0.5}{1-u}+\frac{0.5}{1+u}$$
$$-\frac{0.5}{\sqrt{2}}\,\frac{1}{1-u}+\frac{1}{1+u}$$

I get $$\frac{0.5}{\sqrt{2}}\ln|1-u|-\frac{0.5}{\sqrt{2}}\ln|1+u|+c$$

$$u=\sin\theta=\sqrt{2}\,x$$I get $$\frac{0.5}{\sqrt{2}}\ln|1-\sqrt{2}\,x|-\frac{0.5}{\sqrt{2}}\ln|1+\sqrt{2}\,x|+c$$

$$\frac{\sqrt{2}}{4}\ln|1-\sqrt{2}\,x|-\frac{\sqrt{2}}{4}\ln|1+\sqrt{2}\,x|+c$$
 
  • #7
NotaMathPerson said:


Please check my solution.

$$-\frac{1}{\sqrt{2}}\int\frac{u^2}{1-u^2}\,du$$Using partial fraction decomposition

$$\frac{u^2}{1-u^2}=\frac{u^2}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$$

u=1; A = 0.5
u=-1; B =0.5

$$\frac{0.5}{1-u}+\frac{0.5}{1+u}$$
$$-\frac{0.5}{\sqrt{2}}\,\frac{1}{1-u}+\frac{1}{1+u}$$


No, this should be $$-\frac{0.5}{\sqrt{2}}\left(\frac{1}{1-u}+ \frac{1}{1+u}\right)$$

I get $$\frac{0.5}{\sqrt{2}}\ln|1-u|-\frac{0.5}{\sqrt{2}}\ln|1+u|+c$$

$$u=\sin\theta=\sqrt{2}\,x$$I get $$\frac{0.5}{\sqrt{2}}\ln|1-\sqrt{2}\,x|-\frac{0.5}{\sqrt{2}}\ln|1+\sqrt{2}\,x|+c$$

$$\frac{\sqrt{2}}{4}\ln|1-\sqrt{2}\,x|-\frac{\sqrt{2}}{4}\ln|1+\sqrt{2}\,x|+c$$
 
  • #8
In order for partial fractions to be effective, the degree of the numerator must be strictly less than the degree of the denominator, so you want

$$\frac{u^2}{1-u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1-u)}-1$$

by applying the method outlined by MarkFL.

I'd avoid using decimal numbers in the place of fractions: they may be effective for certain fractions but in general do not always lend themselves to an expression that's easy to work with.
 
  • #9
greg1313 said:
In order for partial fractions to be effective, the degree of the numerator must be strictly less than the degree of the denominator, so you want

$$\frac{u^2}{1-u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1-u)}-1$$

by applying the method outlined by MarkFL.

I'd avoid using decimal numbers in the place of fractions: they may be effective for certain fractions but in general do not always lend themselves to an expression that's easy to work with.

How did you do that expanded form?
 
  • #10
$$\frac{u^2}{1-u^2}=-\frac{1-u^2-1}{1-u^2}=-1+\frac{1}{1-u^2}=-1+\frac12\left(\frac{1}{1+u}+\frac{1}{1-u}\right)$$
 
  • #11
Equivalently, because I am just not as sharp as Greg1313, and have to work it out step by step, I would rewrite the fraction as [tex]\frac{x^2}{1- x^2}= -\frac{x^2}{x^2- 1}[/tex] and now see that both numerator and denominator have degree 2 so divide: [tex]x^2[/tex] goes into [tex]x^2[/tex] once and then there is a remainder of [tex]x^2- (x^2- 1)= 1[/tex]. That is, [tex]\frac{x^2}{1- x^2}= -\frac{x^2}{x^2- 1}= -1- \frac{1}{x^2- 1}[/tex].

Now, I see that [tex]x^2- 1= (x- 1)(x+ 1)[/tex] so, by "partial fractions" that I learned in Calculus, we can write [tex]\frac{1}{x^2- 1}= \frac{A}{x- 1}+ \frac{B}{x+ 1}[/tex] for some constants, A and B. Multiplying both sides by [tex]x^2- 1= (x- 1)(x+ 1)[/tex], [tex]1= \frac{A}{x- 1}(x- 1)(x+ 1)+ \frac{B}{x+ 1}(x- 1)(x+ 1)= A(x+ 1)+ B(x- 1)[/tex].

At this point, there are several ways to determine A and B. Since we want to find two unknown numbers, we need two equations:
1) Multiply on the right to get Ax+ A+ Bx- B= (A+ B)x+ (A- B)= 1= 0x+ 1. In order that two polynomials be the same for all x, the corresponding coefficients must be the same: A+ B= 0, A- B= 1. Adding the two equations, 2A= 1 so A= 1/2 and then 1/2+ B= 0 so B= -1/2.

2) Choose any two values for x to get two equation say x= 0 and x= 2: if x= 0, A(0+ 1)+ B(0- 1)= A- B= 1. If x= 2, A(2+ 1)+ B(2- 1)= 3A+ B= 1. Adding those two equations, 4A= 2 so A= 1/2 and then 1/2- B= 1 so B= 1/2- 1= -1/2.

3) In particular, choose x= 1 and x= -1 since they make the two denominators, x+ 1 and x- 1 equal to 0: if x= 1, A(1+ 1)+ B(1- 1)= 2A= 1 so A= 1/2. If x= -1, A(-1+1)+ B(-1- 1)= -2B= 1 so B= -1/2.

Whichever we use, A= 1/2 and B= -1/2 so [tex]\frac{x^2}{1- x^2}= -1+ \frac{\frac{1}{2}}{x- 1}- \frac{\frac{1}{2}}{x+ 1}[/tex]
 

FAQ: Integration Help 2: Solving Int. w/ Unknown Method

What is integration?

Integration is a mathematical process of finding the area under a curve. It is a way of solving problems involving continuous quantities, such as calculating the distance traveled by a moving object or the amount of water in a tank at a given time.

What is an unknown method in integration?

An unknown method in integration refers to a technique that is not commonly used to solve integration problems. It could be a new approach or a lesser-known method that is not typically taught in math courses.

Why is it important to learn how to solve integration problems using unknown methods?

Learning how to solve integration problems using unknown methods can help expand your problem-solving skills and make you more versatile in approaching mathematical problems. It also allows you to solve more complex and challenging integration problems that may not have a straightforward solution.

What are some common unknown methods used in integration?

Some common unknown methods used in integration include trigonometric substitutions, integration by parts, and partial fractions. Other lesser-known methods include the use of symmetry, the method of undetermined coefficients, and the Laplace transform.

How can I improve my proficiency in solving integration problems with unknown methods?

The best way to improve your proficiency in solving integration problems with unknown methods is to practice regularly and familiarize yourself with different techniques. You can also seek help from online resources or consult with a math tutor for guidance and additional practice problems.

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