If the integral `I=int(x^(5))/(sqrt(1+x^(3)))dx =Ksqrt(x^(3)+1)(x^(3)-2)+C`, (where, C is the constant of integration), then the value of 9K is equal to

Options

A. 4
B. 2
C. 6
D. 10

Answer

To solve the integral \( I = \int \frac{x^5}{\sqrt{1+x^3}} \, dx \) and find the value of \( 9K \) from the given expression \( I = K \sqrt{x^3 + 1}(x^3 - 2) + C \), we will follow these steps:

### Step 1: Substitution
Let \( t = \sqrt{1 + x^3} \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = \frac{3x^2}{2\sqrt{1+x^3}} \, dx \quad \Rightarrow \quad dx = \frac{2t}{3x^2} \, dt \]
From our substitution, we can express \( x^3 \) in terms of \( t \):
\[ x^3 = t^2 - 1 \]

### Step 2: Rewrite the Integral
Now, substitute \( x^5 \) and \( dx \) in the integral:
\[ x^5 = (x^3)^{5/3} = (t^2 - 1)^{5/3} \]
Thus, the integral becomes:
\[ I = \int \frac{(t^2 - 1)^{5/3}}{t} \cdot \frac{2t}{3x^2} \, dt \]
We also need to express \( x^2 \) in terms of \( t \):
\[ x^2 = (t^2 - 1)^{2/3} \]
Now, substituting everything into the integral gives:
\[ I = \int \frac{(t^2 - 1)^{5/3}}{t} \cdot \frac{2t}{3(t^2 - 1)^{2/3}} \, dt = \frac{2}{3} \int (t^2 - 1)^{3/3} \, dt = \frac{2}{3} \int (t^2 - 1) \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ \frac{2}{3} \int (t^2 - 1) \, dt = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + C = \frac{2}{9} t^3 - \frac{2}{3} t + C \]
Substituting back \( t = \sqrt{1 + x^3} \):
\[ I = \frac{2}{9} (1 + x^3)^{3/2} - \frac{2}{3} \sqrt{1 + x^3} + C \]

### Step 4: Compare with Given Expression
Now, we compare this result with the given expression:
\[ K \sqrt{x^3 + 1}(x^3 - 2) + C \]
From our integration, we can see that:
\[ \frac{2}{9} \sqrt{1 + x^3} (x^3 - 2) = K \sqrt{x^3 + 1}(x^3 - 2) \]
Thus, we can equate:
\[ K = \frac{2}{9} \]

### Step 5: Calculate \( 9K \)
Now, we calculate \( 9K \):
\[ 9K = 9 \cdot \frac{2}{9} = 2 \]

### Final Answer
The value of \( 9K \) is \( \boxed{2} \).