This looks like a crazy integral...please help

**s3a** · March 18th 2010, 08:00 PM

How do I evaluate: http://www.wolframalpha.com/input/?i=integral+sqrt((x^2+%2B+1)^2%2F(x^2+-+1)^2)%2C+from+0+to+1%2F2 ?

Any help would be greatly appreciated!
Thanks in advance!

**Prove It** · March 18th 2010, 08:09 PM

Originally Posted by s3a

How do I evaluate: http://www.wolframalpha.com/input/?i=integral+sqrt((x^2+%2B+1)^2%2F(x^2+-+1)^2)%2C+from+0+to+1%2F2 ?

Any help would be greatly appreciated!
Thanks in advance!

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

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

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

$= 1 + \frac{2}{(x - 1)(x + 1)}$ .

Now by using partial fractions:

$\frac{A}{x - 1} + \frac{B}{x + 1} = \frac{2}{(x - 1)(x + 1)}$ .

LHS: $\frac{A}{x - 1} + \frac{B}{x + 1} = \frac{A(x + 1)}{(x - 1)(x + 1)} + \frac{B(x - 1)}{(x - 1)(x + 1)}$

$= \frac{A(x + 1) + B(x - 1)}{(x - 1)(x + 1)}$ .

The numerators are equal, so

$A(x + 1) + B(x - 1) = 2$

$Ax + A + Bx - B = 2$

$(A + B)x + A - B = 0x + 2$ .

So $A + B = 0$ and $A - B = 2$ .

Therefore $A = B + 2$

$B + 2 + B = 0$

$2B + 2 = 0$

$B = -1$ .

$A = B + 2$

$A = -1 + 2$

$A = 1$ .

Therefore $\frac{2}{(x - 1)(x + 1)} = \frac{1}{x - 1} - \frac{1}{x + 1}$ .

So your integral:

$\int{\sqrt{\frac{(x^2 + 1)^2}{(x^2 - 1)^2}}\,dx} = \int{1 + \frac{1}{x - 1} - \frac{1}{x + 1}\,dx}$ .

You should be able to go from here.

**drumist** · March 18th 2010, 09:00 PM

Originally Posted by Prove It

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

Be careful here. Since we are integrating from 0 to 1/2, we have that $x^2-1<0$ , so we should actually get:

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

This doesn't invalidate the rest of the work, but we just need to switch the sign.

In other words, we should instead use:

$\int_0^{1/2} \sqrt{\frac{(x^2 + 1)^2}{(x^2 - 1)^2}}\,dx ~=~ -\int_0^{1/2} \left( 1 + \frac{1}{x - 1} - \frac{1}{x + 1} \right) dx$

**s3a** · March 19th 2010, 10:54 AM

Where does the 1 come from in the integral? (after the square root step)

**Krizalid** · March 19th 2010, 11:05 AM

read above, quite obvious from the previous work shown.

and it's worth to mention that $\frac{2}{(x-1)(x+1)}=\frac{(x+1)-(x-1)}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}.$

**drumist** · March 19th 2010, 11:17 AM

Originally Posted by s3a

Where does the 1 come from in the integral? (after the square root step)

When using partial fraction decomposition, you need the numerator to have lower degree than the denominator. In a more complex problem you may have to use polynomial long division to see the reduction, but in this case it's pretty easy to rewrite the expression to get it.

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

**s3a** · March 19th 2010, 11:52 AM

I still fail to see where the 1 came from. Here is my new work (attached).

**drumist** · March 19th 2010, 12:33 PM

Originally Posted by s3a

I still fail to see where the 1 came from. Here is my new work (attached).

You attempted to use partial fraction decomposition on the following fraction

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

You cannot do that! Partial fraction decomposition cannot be used on this fraction because the degree of the numerator is not less than the degree of the denominator.

Notice how you got

$\frac{1}{x+1} - \frac{1}{x-1}$

when you did the work. But now, let's check the answer:

$\frac{1}{x+1} - \frac{1}{x-1} = \frac{1}{x+1} \left( \frac{x-1}{x-1} \right) - \frac{1}{x-1} \left( \frac{x+1}{x+1} \right)$

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

But this doesn't equal the original fraction.

**s3a** · March 19th 2010, 08:04 PM

I forgot to check the degree of the exponents. After having done long division, etc, I get the right answer! Thanks everyone!

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