Integration Help 2: Solving Int. w/ Unknown Method

[NotaMathPerson]

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?

 

[MarkFL]

Observe that:

$$\frac{2x^2}{2x^2-1}=\frac{2x^2-1+1}{2x^2-1}=1+\frac{1}{2x^2-1}$$

Can you now proceed with a substitution?

 

[HallsofIvy]

There are several different ways to do this but the first thing I would do is multiply both numerator and denominator by -1: $$\frac{-2x^2}{1- 2x^2}dx= -\frac{2x^2}{1- 2x^2}dx$$. I would do that because the denominator, $$1- 2x^2$$, reminds me of a trig substitution. $$sin^2(\theta)+ cos^2(\theta)= 1$$ so $$cos^2(\theta)= 1- sin^2(\theta)$$. If we let $$\sqrt{2}x= sin(\theta)$$ then $$\sqrt{2}dx= cos(\theta)d\theta$$ so $$dx= \frac{1}{\sqrt{2}}cos(\theta)d\theta$$, and [tex]2x^2= sin^2(\theta). <br /> <br /> We have $$\int \frac{2x^2}{1- 2x^2}dx= -\int \frac{2x}{2x^2- 1}dx$$$$= -\int\frac{sin^2(\theta)}{cos^2(\theta)}\left(\frac{1}{\sqrt{2}}cos(\theta)d\theta\right)$$$$= -\frac{1}{\sqrt{2}}\int\frac{sin^2(\theta)}{1- sin^2(\theta)}cos(\theta) d\theta$$ and now the substitution $$u= sin(\theta)$$ changes this to a "rational" function: $$-\frac{1}{\sqrt{2}}\int \frac{u^2}{1- u^2}du$$.[/tex]

 

[NotaMathPerson]

Observe that:

$$\frac{2x^2}{2x^2-1}=\frac{2x^2-1+1}{2x^2-1}=1+\frac{1}{2x^2-1}$$

Can you now proceed with a substitution?

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

$$du=4xdx$$ there's still an x.

 

[MarkFL]

I was thinking more along the lines of a trig. or hyperbolic trig. substitution. You could also try partial fraction decomposition. :)

 

[NotaMathPerson]

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 $$\frac{0.5}{\sqrt{2}}\ln|1-\sqrt{2}\,x|-\frac{0.5}{\sqrt{2}}\ln|1+\sqrt{2}\,x|+c$$

$$\frac{\sqrt{2}}{4}\ln|1-\sqrt{2}\,x|-\frac{\sqrt{2}}{4}\ln|1+\sqrt{2}\,x|+c$$

 

[HallsofIvy]

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 $$\frac{0.5}{\sqrt{2}}\ln|1-\sqrt{2}\,x|-\frac{0.5}{\sqrt{2}}\ln|1+\sqrt{2}\,x|+c$$

$$\frac{\sqrt{2}}{4}\ln|1-\sqrt{2}\,x|-\frac{\sqrt{2}}{4}\ln|1+\sqrt{2}\,x|+c$$

 

[Greg]

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.

 

[NotaMathPerson]

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?

 

[Greg]

$$\frac{u^2}{1-u^2}=-\frac{1-u^2-1}{1-u^2}=-1+\frac{1}{1-u^2}=-1+\frac12\left(\frac{1}{1+u}+\frac{1}{1-u}\right)$$

 

[HallsofIvy]

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 $$\frac{x^2}{1- x^2}= -\frac{x^2}{x^2- 1}$$ and now see that both numerator and denominator have degree 2 so divide: $$x^2$$ goes into $$x^2$$ once and then there is a remainder of $$x^2- (x^2- 1)= 1$$. That is, $$\frac{x^2}{1- x^2}= -\frac{x^2}{x^2- 1}= -1- \frac{1}{x^2- 1}$$.

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

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 $$\frac{x^2}{1- x^2}= -1+ \frac{\frac{1}{2}}{x- 1}- \frac{\frac{1}{2}}{x+ 1}$$