3.5 Q-19

Question Statement

Evaluate the following integral:

$$\int \frac{x}{(x-1)(x^{2}+1)} , dx$$

---

Background and Explanation

To solve this problem, we’ll use the method of partial fraction decomposition. This involves breaking the given rational function into simpler fractions that can be integrated more easily.

We express the integrand as a sum of fractions:

$$\frac{x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$$

where $A$, $B$, and $C$ are constants we need to determine.

---

Solution

We begin by multiplying both sides of the equation by the denominator $(x-1)(x^2+1)$ to eliminate the denominators:

$$x = A(x^2 + 1) + (Bx + C)(x - 1)$$

Next, expand both sides:

$$x = A(x^2 + 1) + (Bx + C)(x - 1)$$ $$x = A(x^2 + 1) + Bx(x - 1) + C(x - 1)$$

Now, let’s expand further:

$$x = A(x^2 + 1) + Bx^2 - Bx + Cx - C$$ $$x = A x^2 + A + Bx^2 - Bx + Cx - C$$

Now, compare the coefficients of like terms on both sides. On the left-hand side, we have just $x$, which means the coefficient of $x^2$ is 0, and the coefficient of $x$ is 1.

Step 1: Coefficients of $x^2$

On the right-hand side, the coefficient of $x^2$ is $A + B$. So, we have:

$$A + B = 0$$

This gives us:

$$B = -A$$

Step 2: Coefficients of $x$

The coefficient of $x$ on the right-hand side is $-B + C$. From the left-hand side, the coefficient of $x$ is 1, so:

$$-B + C = 1$$

Substitute $B = -A$ into this equation:

$$A + C = 1$$

Thus, we get:

$$C = 1 - A$$

Step 3: Constant terms

The constant term on the right-hand side is $A - C$. On the left-hand side, the constant term is 0. Therefore:

$$A - C = 0$$

Substitute $C = 1 - A$ into this:

$$A - (1 - A) = 0$$

Simplifying:

$$A - 1 + A = 0 \quad \Rightarrow \quad 2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}$$

Now that we have $A = \frac{1}{2}$, substitute this back into the equations for $B$ and $C$:

$$B = -A = -\frac{1}{2}, \quad C = 1 - A = 1 - \frac{1}{2} = \frac{1}{2}$$

Thus, the partial fraction decomposition is:

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

Now, we can integrate each term separately:

$$\int \frac{x}{(x-1)(x^2+1)} , dx = \frac{1}{2} \int \frac{1}{x-1} , dx + \frac{1}{2} \int \frac{-x + 1}{x^2+1} , dx$$

We integrate each term:

1. The first term:

$$\frac{1}{2} \int \frac{1}{x-1} , dx = \frac{1}{2} \ln |x-1|$$

2. The second term can be split:

$$\frac{1}{2} \left( \int \frac{-x}{x^2+1} , dx + \int \frac{1}{x^2+1} , dx \right)$$

• The first part, $\int \frac{-x}{x^2+1} , dx$, is a standard integral whose solution is:

$$-\frac{1}{4} \ln(x^2+1)$$

• The second part, $\int \frac{1}{x^2+1} , dx$, is the arctangent function:

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

So the integral becomes:

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

---

Key Formulas or Methods Used

• Partial Fraction Decomposition:

$$\frac{x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$$

• Basic Integrals:
  • $\int \frac{1}{x-1} , dx = \ln |x-1|$
  • $\int \frac{1}{x^2+1} , dx = \tan^{-1}(x)$
  • $\int \frac{-x}{x^2+1} , dx = -\frac{1}{2} \ln(x^2+1)$

---

Summary of Steps

1. Set up Partial Fraction Decomposition: $\frac{x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

2. Multiply through by the denominator: Eliminate denominators by multiplying by $(x-1)(x^2+1)$.

3. Expand and compare coefficients: Solve for $A$, $B$, and $C$ by comparing terms for $x^2$, $x$, and constants.

4. Integrate each term:

   • $\frac{1}{2} \int \frac{1}{x-1} , dx$
   • $\frac{1}{2} \int \frac{-x}{x^2+1} , dx$ and $\frac{1}{2} \int \frac{1}{x^2+1} , dx$

5. Combine the results to get the final solution:

$$\frac{1}{2} \ln |x-1| - \frac{1}{4} \ln(x^2+1) + \frac{1}{2} \tan^{-1}(x) + C$$