`int_(0)^(oo) (dx)/([x+sqrt(x^(2)+1)]^(3))dx=`

Options

A. `(3)/(8)`
B. `(1)/(8)`
C. `-(3)/(8)`
D. none of these

Answer

To solve the integral

\[ I = \int_{0}^{\infty} \frac{dx}{(x + \sqrt{x^2 + 1})^3} \]

we will use a substitution method. Let's go through the solution step by step.

### Step 1: Substitution
Let

\[ t = x + \sqrt{x^2 + 1} \]

Now, we need to express \(x\) in terms of \(t\). Rearranging gives:

\[ \sqrt{x^2 + 1} = t - x \]

Squaring both sides, we get:

\[ x^2 + 1 = (t - x)^2 \]

Expanding the right side:

\[ x^2 + 1 = t^2 - 2tx + x^2 \]

Cancelling \(x^2\) from both sides, we have:

\[ 1 = t^2 - 2tx \]

Rearranging gives:

\[ 2tx = t^2 - 1 \implies x = \frac{t^2 - 1}{2t} \]

### Step 2: Finding \(dx\)
Next, we differentiate \(x\) with respect to \(t\):

\[ dx = \frac{d}{dt}\left(\frac{t^2 - 1}{2t}\right) \]

Using the quotient rule:

\[ dx = \frac{(2t)(2t) - (t^2 - 1)(2)}{(2t)^2} dt = \frac{4t^2 - 2(t^2 - 1)}{4t^2} dt = \frac{4t^2 - 2t^2 + 2}{4t^2} dt = \frac{2t^2 + 2}{4t^2} dt = \frac{t^2 + 1}{2t^2} dt \]

### Step 3: Changing the limits
When \(x = 0\):

\[ t = 0 + \sqrt{0^2 + 1} = 1 \]

When \(x \to \infty\):

\[ t = \infty + \sqrt{\infty^2 + 1} \to \infty \]

Thus, the limits of integration change from \(0\) to \(\infty\) in terms of \(x\) to \(1\) to \(\infty\) in terms of \(t\).

### Step 4: Substitute into the integral
Now substituting \(x\) and \(dx\) into the integral:

\[ I = \int_{1}^{\infty} \frac{1}{t^3} \cdot \frac{t^2 + 1}{2t^2} dt = \frac{1}{2} \int_{1}^{\infty} \frac{t^2 + 1}{t^5} dt \]

This can be separated into two integrals:

\[ I = \frac{1}{2} \left( \int_{1}^{\infty} \frac{1}{t^3} dt + \int_{1}^{\infty} \frac{1}{t^5} dt \right) \]

### Step 5: Evaluate the integrals
Calculating the first integral:

\[ \int_{1}^{\infty} \frac{1}{t^3} dt = \left[-\frac{1}{2t^2}\right]_{1}^{\infty} = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2} \]

Calculating the second integral:

\[ \int_{1}^{\infty} \frac{1}{t^5} dt = \left[-\frac{1}{4t^4}\right]_{1}^{\infty} = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4} \]

### Step 6: Combine the results
Now substituting back into the expression for \(I\):

\[ I = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{4} \right) = \frac{1}{2} \left( \frac{2}{4} + \frac{1}{4} \right) = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8} \]

### Final Result
Thus, the value of the integral is:

\[ \boxed{\frac{3}{8}} \]