prove ∫ 3x² + 1 / [x² - 1]³ dx

To prove the integral of the function \( \frac{3x^2 + 1}{(x^2 - 1)^3} \), we can use the method of partial fractions and integration techniques. Let's break it down step by step.

Step 1: Setting Up the Integral

We start with the integral:

\[ \int \frac{3x^2 + 1}{(x^2 - 1)^3} \, dx \]

First, we can factor the denominator. The expression \( (x^2 - 1) \) can be rewritten as \( (x - 1)(x + 1) \), but since we have it raised to the third power, we will keep it as \( (x^2 - 1)^3 \) for now.

Step 2: Partial Fraction Decomposition

Next, we will express \( \frac{3x^2 + 1}{(x^2 - 1)^3} \) in terms of partial fractions. We can write:

\[ \frac{3x^2 + 1}{(x^2 - 1)^3} = \frac{A}{x^2 - 1} + \frac{B}{(x^2 - 1)^2} + \frac{C}{(x^2 - 1)^3} \]

To find the constants \( A \), \( B \), and \( C \), we multiply through by \( (x^2 - 1)^3 \) to eliminate the denominator:

\[ 3x^2 + 1 = A(x^2 - 1)^2 + B(x^2 - 1) + C \]

Step 3: Expanding and Collecting Terms

Now, we expand the right-hand side:

• For \( A(x^2 - 1)^2 \): \( A(x^4 - 2x^2 + 1) \)
• For \( B(x^2 - 1) \): \( B(x^2 - 1) \)
• And \( C \) remains as \( C \)

Combining these gives:

\[ Ax^4 + (-2A + B)x^2 + (A - B + C) \]

Setting this equal to \( 3x^2 + 1 \), we can equate coefficients:

• For \( x^4 \): \( A = 0 \)
• For \( x^2 \): \( -2A + B = 3 \) (since \( A = 0 \), this simplifies to \( B = 3 \))
• For the constant term: \( A - B + C = 1 \) (substituting \( A = 0 \) and \( B = 3 \) gives \( C = 4 \))

Step 4: Writing the Partial Fraction Decomposition

Now we have:

\[ \frac{3x^2 + 1}{(x^2 - 1)^3} = \frac{3}{(x^2 - 1)^2} + \frac{4}{(x^2 - 1)^3} \]

Step 5: Integrating Each Term

We can now integrate each term separately:

\[ \int \frac{3}{(x^2 - 1)^2} \, dx + \int \frac{4}{(x^2 - 1)^3} \, dx \]

Integrating the First Term

For the first integral, we can use the substitution \( u = x^2 - 1 \), which gives \( du = 2x \, dx \). However, since we don't have an \( x \) in the numerator, we can directly integrate:

\[ \int \frac{3}{(x^2 - 1)^2} \, dx = -\frac{3}{x^2 - 1} + C_1 \]

Integrating the Second Term

For the second integral, we can use a similar approach:

\[ \int \frac{4}{(x^2 - 1)^3} \, dx = -\frac{2}{(x^2 - 1)^2} + C_2 \]

Final Result

Combining both results, we have:

\[ \int \frac{3x^2 + 1}{(x^2 - 1)^3} \, dx = -\frac{3}{x^2 - 1} - \frac{2}{(x^2 - 1)^2} + C \]

Thus, we have successfully proven the integral. If you have any further questions or need clarification on any step, feel free to ask!