Question

Evaluate the indefinite integral:
INT dx/(64+x^2)^2

Please help with this one.

Answer 1

Try substitution
x=8u
dx=8du
x^2=64u^2

I=dx/(64+x^2)=8du/[64(64+64u^2)]
=(1/8)du/(1+u^2)
which is a standard integral.

Answer 2

You can use the method exlained in this posting

http://www.jiskha.com/display.cgi?id=1316565092#1316565092.1316570036


So, we have:

f(x) = 1/(64+x^2)^2

We can factor the denominator as:

(64+x^2)^2 = (x-8 i)^2 (x+8 i)^2

Expand around x = 8 i, by putting:

x = 8 i + t:

f(8i + t) = 1/t^2 1/(16 i + t)^2

We need to expand 1/(16 i + t)^2 to first order in t to get all the singular terms:

1/(16 i + t)^2 =

1/(16 i)^2 1/[1+t/(16i)]^2 =

-1/256 [1 + 8 i t + terms of order t^2 and higher]

So, we have:

f(8i + t) =

-1/256 1/t^2 - i/2048 1/t + non-singular terms

In terms of x, this is:

f(x) = -1/256 1/(x-8 i)^2 -
i/2048 1/(x-8 i) + non-singular terms

Let's call the singular part of thsis expansion s(x):

s(x) = -1/256 1/(x-8 i)^2 -
i/2048 1/(x-8 i)

The singularpart of the expansion around -8i can be found in a similar way, let's call this r(x). Before computing this, consider the ficntion:

g(x) = f(x) - s(x) - r(x)

This is a rational function, but because we've subtracted te singularitoes from f(x) it doesn't have any singularities anymore, therefore it is a polynomial. At infinity, it tends to zero, therefore g(x) is identically zero and we have:

f(x) = s(x) + r(x)

Because f(x) for real x assumes real values, it follows that for real x,
r(x) is the complex conjugate of
s(x). We thus have:

r(x) = -1/256 1/(x+8 i)^2 +
i/2048 1/(x+8 i)

And thus:

f(x) = -1/256 1/(x-8i)^2
-1/256 1/(x+8i)^2 + i/2048 1/(x+8 i) - i/2048 1/(x-8 i)

The first two terms can be directly integrated:

Integral of[ -1/256 1/(x-8i)^2
-1/256 1/(x+8i)^2 ] dx =

1/256 [1/(x-8i) + 1/(x+8i)] =

1/128 x/(x^2 + 64)

The last to terms can be added together before we integrtate to avoid having to deal with complex logarithms:

i/2048 1/(x+8 i) - i/2048 1/(x-8 i) =

i/2048 * (-16 i)/(x^2 + 64) =

1/28 1/(x^2 + 64)

The integral of this is

1/1024 arctan(x/8)

If instead you integrate the last two terms, term by term you get:

i/2048 Log[(x+8 i)/(x-8i)]

Since x is real, the numerator is the complex conjugate of the denominator. The argument of the logarithm thus has a modulus of 1. This means that:

(x+8 i)/(x-8i) = exp(i theta)

for some real theta. You can also see this by writing:

x + 8 i = r exp(i alpha)

Then

x - 8 i = r exp(-i alpha)

The ratio is this

(x+8 i)/(x-8i) = exp(i 2 alpha)

But alpha is, of course, given by:

alpha = arctan(8/x)

So, we have:

Log[(x+8 i)/(x-8i)] =

2 i arctan (8/x)

Multiplying by i/2048 gives the integral:

-1/1024 arctan (8/x) =

1/1024 [arctan(x/8) - pi/2]

The term pi/2 in the brackets can be absorbed into the integration constant.

Answer 3

Thank you Count-Iblis, I did not notice the square in the denominator.
So Carmen, please do not take into account my solution above.