`int_0^(oo) (dx)/([x + sqrt((x^2 + 1))]^3)` =

Options

A. `3//8`
B. `1//8`
C. `-3//8`
D. none

Answer

To solve the integral

\[ I = \int_0^{\infty} \frac{dx}{(x + \sqrt{x^2 + 1})^3} \]

we will use the substitution \( x = \tan \theta \).

### Step 1: Change of Variables
Using the substitution \( x = \tan \theta \), we have:

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

### Step 2: Change the Limits of Integration
When \( x = 0 \), \( \theta = 0 \) and when \( x \to \infty \), \( \theta \to \frac{\pi}{2} \). Thus, the limits change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Rewrite the Integral
Now substituting into the integral, we get:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{\left(\tan \theta + \sqrt{\tan^2 \theta + 1}\right)^3} \]

### Step 4: Simplify the Denominator
We know that \( \sqrt{\tan^2 \theta + 1} = \sec \theta \). Therefore, we can rewrite the denominator:

\[ \tan \theta + \sqrt{\tan^2 \theta + 1} = \tan \theta + \sec \theta \]

Thus, we have:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(\tan \theta + \sec \theta)^3} \]

### Step 5: Further Simplification
Now we express \( \tan \theta \) in terms of sine and cosine:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

So,

\[ \tan \theta + \sec \theta = \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{\sin \theta + 1}{\cos \theta} \]

Thus, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{\left(\frac{\sin \theta + 1}{\cos \theta}\right)^3} = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \cos^3 \theta \, d\theta}{(\sin \theta + 1)^3} \]

### Step 6: Simplifying the Integral
Since \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \), we can simplify:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^3 \theta}{(\sin \theta + 1)^3} \, d\theta \]

### Step 7: Substitution for the Integral
Now, let \( t = 1 + \sin \theta \), then \( d\theta = \frac{dt}{\cos \theta} \) and when \( \theta = 0 \), \( t = 1 \) and when \( \theta = \frac{\pi}{2} \), \( t = 2 \).

### Step 8: Final Integral
The integral becomes:

\[ I = \int_1^2 \frac{\cos^3 \theta}{t^3} \cdot \frac{dt}{\cos \theta} \]

This simplifies to:

\[ I = \int_1^2 \frac{\cos^2 \theta}{t^3} \, dt \]

### Step 9: Evaluating the Integral
Now we can evaluate the integral:

\[ I = \int_1^2 \frac{1}{t^3} \, dt = \left[-\frac{1}{2t^2}\right]_1^2 = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8} \]

### Conclusion
Thus, the value of the integral is:

\[ I = \frac{3}{8} \]