Part 1: Expressing the Integrand as a Sum of Partial Fractions

We are given the integral:

$\int \frac{-10x^2}{x^4 - 1} \, dx$

First, we can factor the denominator:

$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$

So the integrand becomes:

$\frac{-10x^2}{(x - 1)(x + 1)(x^2 + 1)}$

To express this as a sum of partial fractions, we assume:

$\frac{-10x^2}{(x - 1)(x + 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}$

We then multiply both sides by $(x - 1)(x + 1)(x^2 + 1)$ to clear the denominators:

$-10x^2 = A(x + 1)(x^2 + 1) + B(x - 1)(x^2 + 1) + (Cx + D)(x - 1)(x + 1)$

Expanding both sides:

1. For $A(x + 1)(x^2 + 1)$: $A(x + 1)(x^2 + 1) = A(x^3 + x + x + 1) = A(x^3 + 2x + 1)$

2. For $B(x - 1)(x^2 + 1)$: $B(x - 1)(x^2 + 1) = B(x^3 - x + x - 1) = B(x^3 - 2x + 1)$

3. For $(Cx + D)(x - 1)(x + 1)$: $(Cx + D)(x^2 - 1) = Cx(x^2 - 1) + D(x^2 - 1) = Cx^3 - Cx + Dx^2 - D$

Now, combine the terms:

$-10x^2 = A(x^3 + 2x + 1) + B(x^3 - 2x + 1) + (Cx^3 - Cx + Dx^2 - D)$

Group the terms by powers of $x$:

$-10x^2 = (A + B + C)x^3 + (2A - 2B - C)x + (A + B + D)x^2 + (-D)$

Now, equate the coefficients of like powers of $x$:

• Coefficient of $x^3$: $A + B + C = 0$
• Coefficient of $x^2$: $A + B + D = -10$
• Coefficient of $x$: $2A - 2B - C = 0$
• Constant term: $-D = 0$

From the constant term, we get $D = 0$.

Now, solve the system of equations:

1. $A + B + C = 0$
2. $A + B = -10$ (since $D = 0$)
3. $2A - 2B - C = 0$

From equation (2), $A + B = -10$, so substitute $B = -10 - A$ into equations (1) and (3).

• In equation (1): $A + (-10 - A) + C = 0 \implies -10 + C = 0 \implies C = 10$

• In equation (3): $2A - 2(-10 - A) - 10 = 0 \implies 2A + 20 + 2A - 10 = 0 \implies 4A + 10 = 0 \implies A = -\frac{5}{2}$

Now substitute $A = -\frac{5}{2}$ into $A + B = -10$ to find $B$:

$-\frac{5}{2} + B = -10 \implies B = -\frac{15}{2}$

Thus, the partial fraction decomposition is:

$\frac{-10x^2}{(x - 1)(x + 1)(x^2 + 1)} = \frac{-5/2}{x - 1} + \frac{-15/2}{x + 1} + \frac{10x}{x^2 + 1}$

Part 2: Evaluating the Integral

Now we evaluate the integral of each term separately:

$\int \frac{-5/2}{x - 1} \, dx = -\frac{5}{2} \ln|x - 1|$ $\int \frac{-15/2}{x + 1} \, dx = -\frac{15}{2} \ln|x + 1|$ $\int \frac{10x}{x^2 + 1} \, dx = 5 \ln(x^2 + 1)$

Thus, the integral is:

$\int \frac{-10x^2}{x^4 - 1} \, dx = -\frac{5}{2} \ln|x - 1| - \frac{15}{2} \ln|x + 1| + 5 \ln(x^2 + 1) + C$

Would you like more details or clarification on any part of the solution?

Here are some follow-up questions:

1. How do you approach finding the partial fraction decomposition for more complex polynomials?
2. Can you solve integrals of rational functions without partial fractions?
3. What is the significance of the constant term in the solution of partial fractions?
4. How does the choice of substitution affect the simplicity of the integration process?
5. What are some common techniques for evaluating integrals involving logarithms?

Tip: When factoring polynomials in partial fractions, always look for patterns like difference of squares or sum of cubes to simplify your work.