3.5 Q-18

Question Statement

Evaluate the following integral:

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

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Background and Explanation

This is a rational function, which can be solved using the method of partial fraction decomposition. In this method, we express the given function as the sum of simpler fractions whose integrals are easier to evaluate. We will break down the integrand into two parts: one for the factor $(x+1)$ and one for the factor $(x^2+1)$.

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Solution

We begin by setting up the partial fraction decomposition:

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

Step 1: Multiply both sides by $(x+1)(x^2+1)$

Multiplying both sides of the equation by the denominator $(x+1)(x^2+1)$ gives:

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

Step 2: Substitute $x = -1$ to find $A$

Substitute $x = -1$ into equation (ii):

$$-1 - 2 = A((-1)^2 + 1) \quad \Rightarrow 3 = A(2)$$

Thus,

$$A = \frac{-3}{2}$$

Step 3: Expand and compare coefficients

We now expand equation (ii) and compare coefficients. First, expand both sides:

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

Expanding the right-hand side:

$$x - 2 = A x^2 + A + Bx^2 + Bx + Cx + C$$

Simplify:

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

Step 4: Compare coefficients of powers of $x$

• For $x^2$: $A + B = 0$
• For $x$: $B + C = 1$
• For the constant term: $A + C = -2$

Now, substitute $A = -\frac{3}{2}$ into these equations:

• From $A + B = 0$, we have:

$$-\frac{3}{2} + B = 0 \quad \Rightarrow \quad B = \frac{3}{2}$$

• From $B + C = 1$, we substitute $B = \frac{3}{2}$:

$$\frac{3}{2} + C = 1 \quad \Rightarrow \quad C = \frac{-1}{2}$$

Thus, the partial fractions decomposition is:

$$\frac{x-2}{(x+1)(x^2+1)} = \frac{-3}{2(x+1)} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2 + 1}$$

Step 5: Integrate each term

Now, integrate each term separately:

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

This can be split into two integrals:

$$= \frac{-3}{2} \int \frac{1}{x+1} , dx + \frac{3}{2} \int \frac{x}{x^2 + 1} , dx - \frac{1}{2} \int \frac{1}{x^2 + 1} , dx$$

We know the standard integrals:

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

Substituting these into the equation:

$$= \frac{-3}{2} \ln |x+1| + \frac{3}{4} \ln (x^2+1) - \frac{1}{2} \tan^{-1}(x) + C$$

Thus, the final result is:

$$\int \frac{x-2}{(x+1)(x^2+1)} , dx = \frac{-3}{2} \ln |x+1| + \frac{3}{4} \ln (x^2+1) - \frac{1}{2} \tan^{-1}(x) + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Breaking down the rational function into simpler terms.
• Standard Integrals:
  • $\int \frac{1}{x+1} , dx = \ln |x+1|$
  • $\int \frac{x}{x^2+1} , dx = \frac{1}{2} \ln (x^2 + 1)$
  • $\int \frac{1}{x^2+1} , dx = \tan^{-1}(x)$

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Summary of Steps

1. Set up partial fraction decomposition: $\frac{x-2}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+1}$

2. Multiply through by the denominator: $x - 2 = A(x^2 + 1) + (Bx + C)(x + 1)$

3. Solve for $A$, $B$, and $C$ by substituting suitable values for $x$ and comparing coefficients.

4. Integrate each term separately:

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

5. Combine the results to get the final solution: $\frac{-3}{2} \ln |x+1| + \frac{3}{4} \ln (x^2+1) - \frac{1}{2} \tan^{-1}(x) + C$