If `int_(0)^(2) (dx)/( sqrt(x+1) + sqrt((x+1)^(3) ) )` is equal to `kpi` then `k` is equal to

Options

A. `1/2`
B. `2`
C. `1/6`
D. `1/3`

Answer

To solve the integral

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

we will use a substitution method.

### Step 1: Substitution
Let \( t = x + 1 \). Then, we have:

\[ dx = dt \]

### Step 2: Change the limits
When \( x = 0 \), \( t = 0 + 1 = 1 \).
When \( x = 2 \), \( t = 2 + 1 = 3 \).

Thus, the limits of integration change from \( x = 0 \) to \( x = 2 \) into \( t = 1 \) to \( t = 3 \).

### Step 3: Rewrite the integral
Now, substituting \( t \) into the integral, we have:

\[ I = \int_{1}^{3} \frac{dt}{\sqrt{t} + \sqrt{t^3}} \]

### Step 4: Simplify the integrand
We can simplify the integrand:

\[ \sqrt{t^3} = t \sqrt{t} \]

Thus,

\[ I = \int_{1}^{3} \frac{dt}{\sqrt{t} + t \sqrt{t}} = \int_{1}^{3} \frac{dt}{\sqrt{t}(1 + t)} = \int_{1}^{3} \frac{1}{\sqrt{t}(1+t)} dt \]

### Step 5: Further substitution
Let \( u = \sqrt{t} \). Then, \( t = u^2 \) and \( dt = 2u \, du \).

### Step 6: Change the limits again
When \( t = 1 \), \( u = 1 \).
When \( t = 3 \), \( u = \sqrt{3} \).

Thus, the integral becomes:

\[ I = \int_{1}^{\sqrt{3}} \frac{2u \, du}{u(1 + u^2)} = 2 \int_{1}^{\sqrt{3}} \frac{du}{1 + u^2} \]

### Step 7: Evaluate the integral
The integral \( \int \frac{du}{1 + u^2} = \tan^{-1}(u) \). Therefore, we have:

\[ I = 2 \left[ \tan^{-1}(u) \right]_{1}^{\sqrt{3}} = 2 \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \right) \]

### Step 8: Calculate the values
We know that:

\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \quad \text{and} \quad \tan^{-1}(1) = \frac{\pi}{4} \]

Thus,

\[ I = 2 \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = 2 \left( \frac{4\pi - 3\pi}{12} \right) = 2 \left( \frac{\pi}{12} \right) = \frac{\pi}{6} \]

### Step 9: Relate to \( k\pi \)
We are given that \( I = k\pi \). Therefore,

\[ \frac{\pi}{6} = k\pi \]

Dividing both sides by \( \pi \):

\[ k = \frac{1}{6} \]

### Final Answer
Thus, the value of \( k \) is

\[ \boxed{\frac{1}{6}} \]