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

Options

A. (a)`1/2 ln (1+sqrt(1+x^(2)))+C`
B. (b) `2sqrt(1+sqrt(1+x^(2)))+C`
C. (c) `2(1+sqrt(1+x^(2)))+C`
D. (d) None of these

Answer

To solve the integral \( I = \int \frac{x}{\sqrt{1+x^2+\sqrt{(1+x^2)^3}}} \, dx \), we will follow a systematic approach:

### Step 1: Substitution
Let \( t^2 = 1 + x^2 \). Then, differentiating both sides gives:
\[ 2t \, dt = 2x \, dx \implies dt = \frac{x}{t} \, dx \implies x \, dx = t \, dt \]
Now we can rewrite the integral in terms of \( t \).

### Step 2: Rewrite the Integral
Substituting \( x \, dx \) and \( 1 + x^2 \):
\[ I = \int \frac{t \, dt}{\sqrt{t^2 + \sqrt{(t^2)^3}}} \]
Since \( (1+x^2)^3 = (t^2)^3 = t^6 \), we have:
\[ I = \int \frac{t \, dt}{\sqrt{t^2 + t^3}} = \int \frac{t \, dt}{\sqrt{t^2(1+t)}} = \int \frac{t \, dt}{t \sqrt{1+t}} = \int \frac{dt}{\sqrt{1+t}} \]

### Step 3: Further Substitution
Now, let \( v = 1 + t \). Then, \( dt = dv \), and we can rewrite the integral:
\[ I = \int \frac{dv}{\sqrt{v}} \]

### Step 4: Integrate
The integral \( \int v^{-1/2} \, dv \) can be computed as follows:
\[ I = 2v^{1/2} + C = 2\sqrt{v} + C \]

### Step 5: Back Substitute
Now substitute back for \( v \):
\[ v = 1 + t \implies I = 2\sqrt{1+t} + C \]
And since \( t = \sqrt{1+x^2} \):
\[ I = 2\sqrt{1+\sqrt{1+x^2}} + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = 2\sqrt{1+\sqrt{1+x^2}} + C \]