3.5 Q-28

Question Statement

Evaluate the integral:

$$\int \frac{6 a^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+4 a^{2}\right)} , dx$$

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Background and Explanation

This integral involves a rational function with quadratic terms in the denominator. The goal is to decompose the expression into simpler terms using partial fractions. To solve this, we express the given function as a sum of two simpler fractions, each corresponding to one of the quadratic factors in the denominator.

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Solution

Step 1: Set up partial fraction decomposition

We begin by assuming that the integrand can be written as:

$$\frac{6 a^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+4 a^{2}\right)} = \frac{A x + B}{x^{2} + a^{2}} + \frac{C x + D}{x^{2} + 4a^{2}}$$

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Step 2: Multiply both sides by the common denominator

Multiply both sides of the equation by $(x^2 + a^2)(x^2 + 4a^2)$ to eliminate the denominators:

$$6 a^{2} = (A x + B)(x^2 + 4a^2) + (C x + D)(x^2 + a^2)$$

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Step 3: Expand both sides

Next, expand both sides:

$$6 a^2 = A(x^3 + 4a^2 x) + B(x^2 + 4a^2) + C(x^3 + a^2 x) + D(x^2 + a^2)$$

This simplifies to:

$$6 a^2 = (A + C) x^3 + (B + D) x^2 + (4a^2 A + a^2 C) x + (4a^3 B + D a^2)$$

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Step 4: Equate coefficients

Now, we compare the coefficients of like powers of $x$ from both sides of the equation.

• For $x^3$: $A + C = 0$
• For $x^2$: $B + D = 0$
• For $x$: $4 a^2 A + a^2 C = 0$
• For the constant term: $4 a^3 B + D a^2 = 6 a^2$

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Step 5: Solve for constants

From the equation $A + C = 0$, we can conclude that $C = -A$.

Substitute $C = -A$ into the equation $4 a^2 A + a^2 C = 0$:

$$4 a^2 A + a^2 (-A) = 0 \quad \Rightarrow A = 0$$

Since $A = 0$, we also have $C = 0$.

Now, substitute $C = 0$ into the equation $B + D = 0$, yielding $B = -D$.

Substitute into the constant equation $4 a^3 B + D a^2 = 6 a^2$:

$$4 a^3 B + D a^2 = 6 a^2 \quad \Rightarrow 4 a^3 B + (-B) a^2 = 6 a^2$$

This simplifies to:

$$(4 a^3 - a^2) B = 6 a^2 \quad \Rightarrow B = 2$$

Thus, $B = 2$ and $D = -2$.

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Step 6: Rewrite the expression

We can now rewrite the integrand as:

$$\frac{6 a^2}{(x^2 + a^2)(x^2 + 4a^2)} = \frac{2}{x^2 + a^2} - \frac{2}{x^2 + 4a^2}$$

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Step 7: Integrate

Now, we can integrate each term separately:

$$\int \frac{2}{x^2 + a^2} , dx = \frac{2}{a} \tan^{-1}\left(\frac{x}{a}\right)$$ $$\int \frac{2}{x^2 + 4a^2} , dx = \frac{2}{2a} \tan^{-1}\left(\frac{x}{2a}\right)$$

Thus, the final solution is:

$$\int \frac{6 a^2}{(x^2 + a^2)(x^2 + 4a^2)} , dx = \frac{2}{a} \tan^{-1}\left(\frac{x}{a}\right) - \frac{1}{a} \tan^{-1}\left(\frac{x}{2a}\right) + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Used to break down the complex rational function into simpler terms for easier integration.
• Integral of the inverse tangent function:

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

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Summary of Steps

1. Set up partial fraction decomposition: $\frac{6 a^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+4 a^{2}\right)} = \frac{A x + B}{x^{2} + a^{2}} + \frac{C x + D}{x^{2} + 4a^{2}}$
2. Multiply by the common denominator and expand both sides.
3. Equate the coefficients of $x^3$, $x^2$, $x$, and the constant term.
4. Solve the resulting system of equations to find $A = 0$, $B = 2$, $C = 0$, and $D = -2$.
5. Rewrite the integrand as: $\frac{6 a^2}{(x^2 + a^2)(x^2 + 4a^2)} = \frac{2}{x^2 + a^2} - \frac{2}{x^2 + 4a^2}$
6. Integrate each term and combine the results to get the final solution: $\int \frac{6 a^2}{(x^2 + a^2)(x^2 + 4a^2)} , dx = \frac{2}{a} \tan^{-1}\left(\frac{x}{a}\right) - \frac{1}{a} \tan^{-1}\left(\frac{x}{2a}\right) + C$