`int sqrt(x)(1+x^(1//3))^(4)dx " is equal to "`

Options

A. `2{x^(2//3)+(4)/(11)x^(11//6)+(6)/(13)x^(13//6)+(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`
B. `6{x^(2//3)-(4)/(11)x^(11//6)+(6)/(13)x^(13//6)-(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`
C. `6{(1)/(9)x^(3//2)+(4)/(11)x^(11//6)+(6)/(13)x^(13//6)+(4)/(15)x^(5//2)+(1)/(17)x^(17//6)}+c`
D. non of these

Explanation

To solve the integral \(\int \sqrt{x} (1 + x^{1/3})^4 \, dx\), we will follow these steps:

### Step 1: Substitution
Let \(x = t^6\). Then, we have:
\[ dx = 6t^5 \, dt \]
Substituting these into the integral gives:
\[ \int \sqrt{t^6} (1 + (t^6)^{1/3})^4 \cdot 6t^5 \, dt \]
This simplifies to:
\[ \int t^3 (1 + t^2)^4 \cdot 6t^5 \, dt \]

### Step 2: Simplifying the Integral
Now we can combine the powers of \(t\):
\[ 6 \int t^8 (1 + t^2)^4 \, dt \]

### Step 3: Expanding the Binomial
Next, we expand \((1 + t^2)^4\) using the binomial theorem:
\[ (1 + t^2)^4 = \sum_{k=0}^{4} \binom{4}{k} (t^2)^k = 1 + 4t^2 + 6t^4 + 4t^6 + t^8 \]
Thus, we have:
\[ 6 \int t^8 (1 + 4t^2 + 6t^4 + 4t^6 + t^8) \, dt \]

### Step 4: Distributing \(t^8\)
Distributing \(t^8\) gives:
\[ 6 \int (t^8 + 4t^{10} + 6t^{12} + 4t^{14} + t^{16}) \, dt \]

### Step 5: Integrating Each Term
Now we can integrate each term:
\[ 6 \left( \frac{t^9}{9} + \frac{4t^{11}}{11} + \frac{6t^{13}}{13} + \frac{4t^{15}}{15} + \frac{t^{17}}{17} \right) + C \]

### Step 6: Substituting Back
Now we substitute back \(t = x^{1/6}\):
\[ = 6 \left( \frac{x^{3/2}}{9} + \frac{4(x^{11/6})}{11} + \frac{6(x^{13/6})}{13} + \frac{4(x^{15/6})}{15} + \frac{x^{17/6}}{17} \right) + C \]

### Final Answer
Thus, the final answer is:
\[ \frac{2}{3} x^{3/2} + \frac{24}{11} x^{11/6} + \frac{36}{13} x^{13/6} + \frac{24}{15} x^{15/6} + \frac{6}{17} x^{17/6} + C \]