integral of 3/1-3x^2
To find the integral of the function 3 / (1 - 3x^2), we can make use of the technique of partial fractions.
Step 1: Factorize the denominator
The denominator in our expression is 1 - 3x^2. We can factorize it as follows:
1 - 3x^2 = (1 - √3x)(1 + √3x)
Step 2: Express the fraction in partial fractions
We can express the original fraction as a sum of two partial fractions:
3 / (1 - 3x^2) = A / (1 - √3x) + B / (1 + √3x)
Step 3: Find the values of A and B
To find the values of A and B, we can multiply both sides of the equation by the denominator:
3 = A(1 + √3x) + B(1 - √3x)
Now, let's solve for A and B:
Setting x = -1/√3, we get:
3 = A(1 + √3(-1/√3)) + B(1 - √3(-1/√3))
3 = A(1 - 1) + B(1 + 1)
3 = 0 + 2B
B = 3/2
Next, setting x = 1/√3, we get:
3 = A(1 + √3(1/√3)) + B(1 - √3(1/√3))
3 = A(1 + 1) + B(1 - 1)
3 = 2A + 0
A = 3/2
So, we have A = 3/2 and B = 3/2.
Step 4: Evaluate the integral
Now that we have our partial fractions, we can integrate each term separately:
∫ (3 / (1 - 3x^2)) dx = ∫ (3/2) / (1 - √3x) dx + ∫ (3/2) / (1 + √3x) dx
Using the formula for the integral of 1/u, we can evaluate each integral:
∫ (3/2) / (1 - √3x) dx = (3/2) * ln|1 - √3x| + C1
∫ (3/2) / (1 + √3x) dx = (3/2) * ln|1 + √3x| + C2
So, the final result is:
∫ (3 / (1 - 3x^2)) dx = (3/2) * ln|1 - √3x| + C1 + (3/2) * ln|1 + √3x| + C2
Note: C1 and C2 are constants of integration and can be combined into a single constant if desired.