Partial fractions & Substitution Integration

[thomas49th]

Homework Statement

Hi,

\int \frac{1}{x(x^{2}+1)}dx

Homework Equations

The Attempt at a Solution

well I split this into partial fractions

\frac{A}{x} + \frac{Bx + C}{x^{2} + 1}<br /> <br /> so 1 \equiv A(x^{2}+1) + (Bx + C)x<br /> <br /> when x = 0, A =1 <br /> <br /> when x = 1, Bx + C = -1 so comparing coeffs tell us B = 0 and C = -1 <br /> Correct?<br /> <br /> Now integrating it:<br /> <br /> \int \frac{1}{x} - \frac{1}{x^{2} + 1}dx<br /> <br /> the first one is just ln |x| <br /> <br /> and the second one<br /> <br /> let u = x^{2} + 1<br /> <br /> u&#039; = 2x<br /> <br /> so<br /> <br /> ln|x| - \int \frac{1}{2x u}du<br /> <br /> Have I gone wrong somewhere. <br /> <br /> Thanks<br /> Thomas

 

[Count Iblis]

Yoiu have to express all the x's in terms of u after you do the substitution. But there is another error that you can immediately spot.

The function 1/[x(x^2 + 1) ]

clearly goes to zero like x^(-3) for large x. But what you obtained for the partial fraction expansion is not consistent with this. In fact, you can obtain the partial fraction expansion simply by studying the behavior of the function without solving any equations, as follows.

Clearly the singular behavior near zero must be given as:

1/x

Then to get the correct behaviour at infinity, we need to cancel the 1/x behavaviour from this tarm at infinity. To do that, we need need a term:

-x/(x^2 + 1)

Is there room for an A/(x^2 + 1) term?

No, because the function has to be odd and the even part due to this contribution cannot be canceled by another term in the partial fraction exapansion as there are no more terms to consider.

 

[thomas49th]

I can see what you're saying in the first part. I have plotted the function and then plotted my partial fractions on a graph and I can see they don't lie on top of one another => someone made an error.

But

Clearly the singular behavior near zero must be given as:

1/x

Then to get the correct behaviour at infinity, we need to cancel the 1/x behavaviour from this tarm at infinity. To do that, we need need a term:

-x/(x^2 + 1)

Is there room for an A/(x^2 + 1) term?

No, because the function has to be odd and the even part due to this contribution cannot be canceled by another term in the partial fraction exapansion as there are no more terms to consider.

Makes little sense to me. I know that as x tends to infinity 1/x tends to 0. What do you mean when you say "singular". Do you mean infinity? I know of odd and even functions and I understand that 1/(x^2 + 1) is an even function, but 1/x is an odd because it remains unchanged after 180°, but why is this important?

Thanks
Thomas

 

[HallsofIvy]

Homework Statement

Hi,

\int \frac{1}{x(x^{2}+1)}dx

Homework Equations

The Attempt at a Solution

well I split this into partial fractions

\frac{A}{x} + \frac{Bx + C}{x^{2} + 1}<br /> <br /> so 1 \equiv A(x^{2}+1) + (Bx + C)x<br /> <br /> when x = 0, A =1 <br /> <br /> when x = 1, Bx + C = -1 so comparing coeffs tell us B = 0 and C = -1 <br /> Correct?

<br /> No. <br /> \frac{1}{x}- \frac{1}{x^2+ 1}= \frac{x^2+ 1- x}{x(x^2+ 1)}\ne \frac{1}{x(x^2+1)}<br /> <br /> You cannot say &quot;when x= 1, Bx+ C= -1&quot;, because with x=1 you do not have an &quot;x&quot; in the formula. You have B+ C= 1 which is not sufficient to determine both B and C.<br /> <br /> You <b>can</b> say &quot;with A= 1, this become 1= x^2+ 1+ Bx^2+ C so that Bx^2+ C= 1- x^2- 1= -x^2 and now B= -1 and C= 0.&quot;<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Now integrating it:<br /> <br /> \int \frac{1}{x} - \frac{1}{x^{2} + 1}dx<br /> <br /> the first one is just ln |x| <br /> <br /> and the second one<br /> <br /> let u = x^{2} + 1<br /> <br /> u&#039; = 2x<br /> <br /> so<br /> <br /> ln|x| - \int \frac{1}{2x u}du </div> </div> </blockquote> Oh, dear, oh dear! When you substitute &quot;u&quot; for some function of x you cannot have both &quot;u&quot; and &quot;x&quot; in the integral because you cannot then integrate with respect to u treating x as a constant! You cannot make this substitution be cause you do not already have an &quot;x&quot; in the numerator to give &quot;xdx&quot;.<br /> <br /> What you can do here is recognize that<br /> \int \frac{dx}{x^2+ 1}= arctan x+ C<br /> <br /> Of course, since you have the values of B and C reversed, you don&#039;t have that integral at all- you have <br /> \int \frac{xdx}{x^2+ 1} <br /> and you CAN use that substitution.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Have I gone wrong somewhere. <br /> <br /> Thanks<br /> Thomas </div> </div> </blockquote>

 

[rl.bhat]

You can try this method.
Multiply by x to numerator and denominator. The problem becomes
Intg[x*dx/x^2(x^2 + 1)]
Now substitute x^2 = u, then 2x*dx = du or x*dx = du/2
The problem becomes
Intg[du/2u(u+1)]
Now find the partial factors and solve by taking integration.