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101 Dec 21, 2007 at 16:05

Having taken Calculus over 35 years ago and did pretty well, I am now a bit stumped in solving this. I have factered 1/4 to the left of the integrand, and let u=2x\^3+1, and then get du/dx=6x\^2, then substitute back in but get stumped at that point.

Really would appreciate some help on this one.

Byron

#### 8 Replies

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101 Dec 21, 2007 at 18:36

Hi Byron,

My symbolic math software tells me that the integral of that equation is:

LN((2·x\^2 + 1)·ABS(x))/4

(strange! I would have expected a rational polyonom..)

Btw - I use “derive for windows” from texas instruments for such math things.. It is not as powerfull as mathlab but much easier to learn and work with.

TI has a demo of it on their webpage. Give it a try.

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101 Dec 21, 2007 at 19:07

That is not the answer that I got over at http://integrals.wolfram.com/index.jsp That site says it evaluates to:
1/4logx + log(2x\^2+1)

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101 Dec 21, 2007 at 19:12

Hm. These equations are related but not identical.

Just wondering: You you need the integral of:

((6x\^2+1)/(8x\^3+4x))

of the integral of:

((6x\^2+1)/(8x\^3+4x))dx

Nils

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101 Dec 21, 2007 at 19:26

@bosshog

That is not the answer that I got over at http://integrals.wolfram.com/index.jsp That site says it evaluates to:
1/4logx + log(2x\^2+1)

Hm - just thinking:

((6x\^2+1)/(8x\^3+4x)) is well defined for any x except 0 (because the divisor-polynom has a root at zero).

The wolfram integrator evaluates this to:

1/4logx + log(2x\^2+1)

And this is only defined for positive x values since log(x) | x<=0 is undefined as long as we’re working with real values. So the integral from the wolfram site must be flawed.

The integral from derive:

LN((2·x\^2 + 1)·ABS(x))/4

Is symetrical at x=0 and well-defined over the entire range. That i what I expect when I plot the integral and the equation we started with.

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101 Dec 21, 2007 at 20:43

Hi,

Not sure if this is right, but here is an attempt to get it directly. If you start from your original integral,

int( (6x\^2+1)/(8x\^3+4x) )

now, factor out the 1/4 and rearrange the bottom slightly:

1/4 * int( (6x\^2+1)/(x(2x\^2+1)) )

and rewrite the top part:

1/4 * int( (4x\^2 + 2x\^2+1) / (x(2x\^2+1)) )

we can now cancel some parts and we get:

1/4 * ( int( 4x / (2x\^2+1) ) + int( 1/x ) )
1/4 * ( int( 4x / (2x\^2+1) ) + ln(|x|) )
1/4 * ( ln( |2x\^2+1| ) + ln(|x|) )

I may very well have made an error somewhere though :(

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165 Dec 21, 2007 at 21:05

That answer is identical to the one Nils got, if you combine the logs (note the abs bars around 2x\^2 + 1 are unnecessary since 2x\^2 + 1 is always positive anyways).

The Wolfram site seems to incorrectly drop the abs bars from the ‘x’ term though.

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101 Dec 21, 2007 at 22:46

I get the same results when I do it on paper.

.oO(maybe we should get in touch with the wolfram guys - we can be wrong but it seems like they have a bug in their online-integrator).

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101 Dec 22, 2007 at 03:26

With TI89-Titanium i get:

 (ln(((sqrt(2)*x-1)\^2 * (sqrt(2)*x+1)\^2)/( x ))) / (4)

which looks like this: