Integral of 1/[x^2(1+x^2)]
To find the integral of 1/[x^2(1+x^2)], we can use a technique called partial fractions. This technique allows us to break down the given expression into simpler fractions.
Step 1: Factorize the denominator:
The denominator can be factored as (x^2)(1 + x^2).
Step 2: Write the partial fraction decomposition:
Since the denominator has two distinct linear factors, we can write the given expression as a sum of two fractions with simpler denominators. Let's write it as:
1/[x^2(1 + x^2)] = A/x^2 + B/(1 + x^2).
Step 3: Find the values of A and B:
To determine the values of A and B, we need to find a common denominator for the right-hand side. Multiplying through by x^2(1 + x^2), we get:
1 = A(1 + x^2) + B(x^2).
Next, we can substitute specific values for x that will help us solve for A and B.
For x = 0:
1 = A(1) + B(0).
1 = A.
For x = ±i (the imaginary unit):
1 = A(1 - 1) + B(-1).
1 = -B.
B = -1.
So, we have A = 1 and B = -1.
Step 4: Substitute the values back into the original partial fraction decomposition:
1/[x^2(1 + x^2)] = 1/x^2 - 1/(1 + x^2).
Step 5: Integrate each term:
∫(1/x^2 - 1/(1 + x^2)) dx = ∫(1/x^2) dx - ∫(1/(1 + x^2)) dx.
The integral of 1/x^2 is -1/x.
Now, let's focus on the second term:
∫(1/(1 + x^2)) dx.
To integrate this term, we can make use of the substitution method. Let u = 1 + x^2, and therefore, du/dx = 2x, or dx = du/(2x).
Substituting du/(2x) for dx in the integral, we get:
∫(1/(1 + x^2)) dx = ∫(1/u) * (du/(2x)).
Simplifying this:
∫(1/(1 + x^2)) dx = (1/2) ∫(1/u) du.
The integral of 1/u is ln|u|, so:
∫(1/(1 + x^2)) dx = (1/2) ln|u| + C.
Substituting back the value of u = 1 + x^2:
∫(1/(1 + x^2)) dx = (1/2) ln|1 + x^2| + C.
Therefore, the integral of 1/[x^2(1 + x^2)] is:
∫(1/[x^2(1 + x^2)]) dx = ∫(1/x^2) dx - ∫(1/(1 + x^2)) dx
= -1/x + (1/2) ln|1 + x^2| + C.
Note: C represents the constant of integration.