`int _(0)^(a) (x^(4)dx)/((a^(2)+x^(2))^(4))` is equal to

Options

A. `(1)/(16a^(3))((pi)/(4)-(1)/(3))`
B. `(1)/(16a^(3))((pi)/(4)+(1)/(3))`
C. `(1)/(16)a^(3)((pi)/(4)-(1)/(3))`
D. `(1)/(16)a^(3)((pi)/(4)+(1)/(3))`

Explanation

To solve the integral

\[ I = \int_{0}^{a} \frac{x^4 \, dx}{(a^2 + x^2)^4} \]

we will use the substitution method. Let's go through the steps:

### Step 1: Substitution
Let \( x = a \tan \theta \). Then, we have:

\[ dx = a \sec^2 \theta \, d\theta \]

When \( x = 0 \), \( \theta = 0 \) and when \( x = a \), \( \theta = \frac{\pi}{4} \).

### Step 2: Change the limits and substitute into the integral
Substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{(a \tan \theta)^4 \cdot a \sec^2 \theta \, d\theta}{(a^2 + (a \tan \theta)^2)^4} \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{a^5 \tan^4 \theta \sec^2 \theta \, d\theta}{(a^2(1 + \tan^2 \theta))^4} \]

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{a^5 \tan^4 \theta \sec^2 \theta \, d\theta}{(a^2 \sec^2 \theta)^4} = \int_{0}^{\frac{\pi}{4}} \frac{a^5 \tan^4 \theta \sec^2 \theta \, d\theta}{a^8 \sec^8 \theta} \]

### Step 3: Simplifying the integral
This simplifies to:

\[ I = \frac{1}{a^3} \int_{0}^{\frac{\pi}{4}} \tan^4 \theta \sec^{-6} \theta \, d\theta \]

### Step 4: Rewrite in terms of sine and cosine
Using \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \):

\[ I = \frac{1}{a^3} \int_{0}^{\frac{\pi}{4}} \frac{\sin^4 \theta}{\cos^4 \theta} \cdot \cos^6 \theta \, d\theta = \frac{1}{a^3} \int_{0}^{\frac{\pi}{4}} \sin^4 \theta \cos^2 \theta \, d\theta \]

### Step 5: Use the identity for sine
Using the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):

\[ \sin^4 \theta = \left(\sin^2 \theta\right)^2 = \left(\frac{1 - \cos 2\theta}{2}\right)^2 = \frac{1 - 2\cos 2\theta + \cos^2 2\theta}{4} \]

### Step 6: Substitute and integrate
Substituting this back into the integral:

\[ I = \frac{1}{a^3} \int_{0}^{\frac{\pi}{4}} \left(\frac{1 - 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}}{4}\right) \cos^2 \theta \, d\theta \]

Now, we can integrate term by term.

### Step 7: Evaluate the integral
This integral can be evaluated using standard techniques, leading to:

\[ I = \frac{1}{16 a^3} \left( \frac{\pi}{4} - \frac{1}{3} \right) \]

### Final Result
Thus, the final result is:

\[ I = \frac{\pi}{64 a^3} - \frac{1}{48 a^3} \]