Integral of 1/[x^2(1+x^2)]
I would do this with partial fractions.
Please do,I'm waiting...
rather than just sit around waiting, why not actually try it yourself?
1/[x^2(1+x^2)] = 1/x^2 - 1/(1+x^2)
and those are both easy, right?
To find the integral of 1/[x^2(1+x^2)], we can start by attempting to simplify the integrand. We notice that the denominator can be written as (x^2) * (1+x^2).
Now, let's try to factorize this expression further.
We can rewrite (1+x^2) as (x^2 + 1).
Therefore, the denominator becomes x^2 * (x^2 + 1).
Now, we can rewrite the integral as ∫ (1/[x^2 * (x^2 + 1)]) dx.
To proceed, we can use a method called Partial Fraction Decomposition to break down the integrand into simpler fractions.
We assume that A, B, C, and D are constants and express the integrand as follows:
1/[x^2 * (x^2 + 1)] = A/x + B/x^2 + C/(x+i) + D/(x-i).
Next, we find a common denominator by multiplying both sides of the equation by x^2 * (x^2 + 1).
1 = A * (x^2 * (x+i) * (x-i)) + B * ((x^2 + 1) * (x+i) * (x-i)) + C * (x * (x^2 + 1) * (x-i)) + D * (x * (x^2 + 1) * (x+i)).
After expanding and simplifying, we get:
1 = A * (x^4 - x^2) + B * (x^4 + x^3 + x^2 - x) + C * (x^4 + x^2) + D * (x^4 + x^2).
Now, we can collect the terms containing x^4, x^3, x^2, x, and the constants.
1 = (A + B + C + D) * x^4 + (C + D) * x^2 + (- B) * x + (A).
Equating the coefficients on both sides, we get the following set of equations:
A + B + C + D = 0,
C + D = 0,
- B = 0,
A = 1.
Solving these equations, we find A = 1, B = 0, C = 0, and D = -1.
Thus, the integral becomes:
∫ (1/[x^2 * (x^2 + 1)]) dx = ∫ (1/x - 1/[x^2 + 1]) dx = ∫ (1/x) dx - ∫ (1/[x^2 + 1]) dx.
Integrating each term separately, we get:
= ln|x| - arctan(x) + C,
where C is the constant of integration.
Therefore, the integral of 1/[x^2(1+x^2)] is ln|x| - arctan(x) + C.