Evaluate the indefinite integral:
INT dx/(64+x^2)^2
Please help with this one.
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.
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.
Thank you Count-Iblis, I did not notice the square in the denominator.
So Carmen, please do not take into account my solution above.
To evaluate the indefinite integral of dx/(64+x^2)^2, we can use a substitution.
Start by letting u = x^2. Then, du/dx = 2x, which implies dx = (1/2x)du.
Now, substitute the expressions for x and dx into the original integral:
∫ dx/(64+x^2)^2 = ∫ (1/2x)du/(64+u)^2
Next, express x^2 in terms of u: x^2 = u.
Therefore, the integral becomes:
∫ (1/2x)du/(64+x^2)^2 = ∫ (1/2√u)(1/u)/(64+u)^2.
Combine the terms with u in the denominator:
∫ (1/(2u√u))/(64+u)^2.
Now, we can simplify the integral by splitting it into two separate fractions:
∫ 1/(2u(64+u)^2) * 1/√u du.
Separate the two fractions:
(1/2) ∫ 1/(u(64+u)^2) du * ∫ 1/√u du.
Both of these integrals can be evaluated using basic integration rules.
For the first integral, let's use partial fraction decomposition. We can write the integrand as the sum of two fractions:
1/(u(64+u)^2) = A/u + B/(64+u) + C/(64+u)^2.
To find the values of A, B, and C, we need to find a common denominator and equate the numerator coefficients:
1 = A(64+u)^2 + Bu(64+u) + Cu.
Now, we can solve for A, B, and C.
The second integral, ∫ 1/√u du, is a simple power rule integration:
∫ 1/√u du = 2√u.
Now that we have the values of A, B, and C, and the other integral result, we can substitute them back in to evaluate the original integral.