I think that the key here is to make the substitution u = (x+1)^1/2 . This makes the problem the integral of:

2/(u⁴ - 2u² + 4)

The denominator can be split into:

(u-Ö2e^ip/12)(u+Ö2e^ip/12)(u-Ö2e^-ip/12) (u+Ö2e^-ip/12)
So you can split the fraction into partial fractions of the form:

A/(x-p) + B/(x-q) + C/(x-r) + D/(x-s)

which can be integrated.

This is not a particularly nice method - does anyone have any better ideas?

Michael

Michael,

The denominator can be factorised as

[u² +sqrt(6)u+2][u² -sqrt(6)u+2]

so you can use partial fraction - notice that the denominator will be of the form (v² +a² ). This saves the trouble of using complex numbers.

Is this method any nicer?

Kerwin

Yes, sorry - that's much nicer. With your factorisation at least we are going to avoid having to spend ages separating the arctans from the lns. If I get a bit of time I will post the final answer soon.

Thanks for that,

Michael

By the way, it is interesting to note that if you combine the two methods of obtaining the same integral you can prove all the rules about complex numbers. With a simpler example let

dI/dx = 1/(x² +1) (I = 0 when x = 0)

then this can be written as

dI/dx = i/2[1/(x+i)-1/(x-i)]

So I = arctan(x) and I = i/2ln((i+x)/(i-x))

Therefore arctan(x) = i/2ln((i+x)/(i-x))

Then let x tend to infinity to see why e^pi=-1! (I know that last equation is obvious anyway but this is a good way of proving it beyond doubt.)

Michael

Okay, back to the problem.

We got to

I = integral[2/(u⁴ -2u² +4)]du

Using Kerwin's factorisation we can write the integrand as:

(1/(2sqrt(6))u + 1/2)/(u² +sqrt(6)u+2) + (-1/(2sqrt(6))u + 1/2)/(u² -sqrt(6)u+2)

Making the integral:

1/(4sqrt(6)) ln(u² +sqrt(6)u+2) - 1/(4sqrt(6)) ln(u² -sqrt(6)u+2) + 1/(2sqrt(2)) arctan(sqrt(2)u+sqrt(3)) + 1/(2sqrt(2))arctan(sqrt(2)u-sqrt(3))

Substituting u = sqrt(x+1) back to get:

1/(4sqrt(6)) ln(x+3+sqrt(6x+6)) - 1/(4sqrt(6)) ln(x + 3 - sqrt(6x + 6)) + 1/(2sqrt(2)) arctan(sqrt(2x+2)+sqrt(3)) + 1/(2sqrt(2)) arctan(sqrt(2x+2)-sqrt(3)) + C

(I think! I would be surprised if I got through all that without an error.)

Michael