`int ( x ^(3))/( (x ^(2) + 1) (x ^(2) + 4 ))dx` is equal to

Options

A. `tan ^(-1) x +2 tan ^(-1) "" ((x)/(2)) + C`
B. `tan ^(-1) ((x)/(2)) - 4 tan ^(-1) x +C`
C. `-1/2 tan ^(-1) x + (2)/(3) tan ^(-1) (( x )/(2)) +C`
D. `4 tan ^(-1) ((x)/(2)) - 2 tan ^(-1) x +C`

Answer

To solve the integral

\[ I = \int \frac{x^3}{(x^2 + 1)(x^2 + 4)} \, dx, \]

we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form:

\[ I = \int \frac{x^3}{(x^2 + 1)(x^2 + 4)} \, dx. \]

### Step 2: Use Substitution
We can use the substitution \( t = x^2 \). Then, \( dt = 2x \, dx \) or \( dx = \frac{dt}{2\sqrt{t}} \).

Substituting \( x^2 = t \) into the integral gives:

\[ I = \int \frac{t^{3/2}}{(t + 1)(t + 4)} \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int \frac{t}{(t + 1)(t + 4)} \, dt. \]

### Step 3: Simplify the Integral
Now, we simplify the integral:

\[ I = \frac{1}{2} \int \frac{t}{(t + 1)(t + 4)} \, dt. \]

### Step 4: Perform Partial Fraction Decomposition
Next, we perform partial fraction decomposition on the integrand:

\[ \frac{t}{(t + 1)(t + 4)} = \frac{A}{t + 1} + \frac{B}{t + 4}. \]

Multiplying through by the denominator \((t + 1)(t + 4)\) gives:

\[ t = A(t + 4) + B(t + 1). \]

### Step 5: Solve for A and B
Expanding and collecting like terms, we have:

\[ t = (A + B)t + (4A + B). \]

By comparing coefficients, we get:

1. \( A + B = 1 \)
2. \( 4A + B = 0 \)

From the first equation, \( B = 1 - A \). Substituting into the second equation:

\[ 4A + (1 - A) = 0 \implies 3A + 1 = 0 \implies A = -\frac{1}{3}. \]

Then, substituting back to find \( B \):

\[ B = 1 - \left(-\frac{1}{3}\right) = \frac{4}{3}. \]

### Step 6: Rewrite the Integral
Now we can rewrite the integral using these values:

\[ I = \frac{1}{2} \int \left(-\frac{1}{3(t + 1)} + \frac{4}{3(t + 4)}\right) dt. \]

### Step 7: Integrate
Integrating term by term:

\[ I = \frac{1}{2} \left(-\frac{1}{3} \ln |t + 1| + \frac{4}{3} \ln |t + 4| \right) + C. \]

### Step 8: Substitute Back
Substituting \( t = x^2 \) back into the equation gives:

\[ I = \frac{1}{2} \left(-\frac{1}{3} \ln |x^2 + 1| + \frac{4}{3} \ln |x^2 + 4| \right) + C. \]

### Step 9: Simplify the Final Expression
This simplifies to:

\[ I = -\frac{1}{6} \ln(x^2 + 1) + \frac{2}{3} \ln(x^2 + 4) + C. \]

### Final Answer
Thus, the final result for the integral is:

\[ I = -\frac{1}{6} \ln(x^2 + 1) + \frac{2}{3} \ln(x^2 + 4) + C. \]