Find `int(dx)/((x^(2)+a^(2))(x^(2)+b^(2)))` equals

Options

A. `(1)/(b^(2)-a^(2))[(1)/(b)tan^(-1)((x)/(a))-(1)/(a) tan^(-1)(x)]+c`
B. `(1)/(b^(2)-a^(2))[(1)/(a)tan^(-1)((x)/(a))-(1)/(b)tan^(-1)((x)/(b))]+c`
C. `(1)/(a)tan^(-1)((x)/(a))+(1)/(b)tan^(-1)((x)/(b))+c`
D. `(1)/(a)tan^(-1)((x)/(a))-(1)/(b)tan^(-1)((x)/(b))+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{(x^2 + a^2)(x^2 + b^2)}, \]

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition

We can express the integrand as:

\[ \frac{1}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}, \]

where \(A\) and \(B\) are constants to be determined.

### Step 2: Multiply through by the denominator

Multiplying both sides by \((x^2 + a^2)(x^2 + b^2)\) gives:

\[ 1 = A(x^2 + b^2) + B(x^2 + a^2). \]

### Step 3: Expand and collect like terms

Expanding the right side, we have:

\[ 1 = Ax^2 + Ab^2 + Bx^2 + Ba^2 = (A + B)x^2 + (Ab^2 + Ba^2). \]

### Step 4: Set up equations for coefficients

For the equation to hold for all \(x\), the coefficients of \(x^2\) must be equal to zero, and the constant terms must equal one:

1. \(A + B = 0\)
2. \(Ab^2 + Ba^2 = 1\)

### Step 5: Solve the system of equations

From the first equation, we can express \(B\) in terms of \(A\):

\[ B = -A. \]

Substituting \(B\) into the second equation gives:

\[ A b^2 - A a^2 = 1 \implies A(b^2 - a^2) = 1 \implies A = \frac{1}{b^2 - a^2}. \]

Then,

\[ B = -\frac{1}{b^2 - a^2}. \]

### Step 6: Rewrite the integral

Now substituting \(A\) and \(B\) back into the partial fractions gives:

\[ \frac{1}{(x^2 + a^2)(x^2 + b^2)} = \frac{1}{b^2 - a^2} \left( \frac{1}{x^2 + a^2} - \frac{1}{x^2 + b^2} \right). \]

Thus, we rewrite the integral \(I\):

\[ I = \int \frac{1}{(x^2 + a^2)(x^2 + b^2)} \, dx = \frac{1}{b^2 - a^2} \left( \int \frac{1}{x^2 + a^2} \, dx - \int \frac{1}{x^2 + b^2} \, dx \right). \]

### Step 7: Evaluate the integrals

We know that:

\[ \int \frac{1}{x^2 + c^2} \, dx = \frac{1}{c} \tan^{-1} \left( \frac{x}{c} \right) + C. \]

Thus, we have:

\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C, \]
\[ \int \frac{1}{x^2 + b^2} \, dx = \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) + C. \]

### Step 8: Substitute back into the integral

Substituting these results back, we get:

\[ I = \frac{1}{b^2 - a^2} \left( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) - \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) \right) + C. \]

### Final Result

Thus, the final result for the integral is:

\[ I = \frac{1}{b^2 - a^2} \left( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) - \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) \right) + C. \]