If `int x^(26).(x-1)^(17).(5x-3)dx=(x^(27).(x-1)^(18))/(k)+C` where C is a constant of integration, then the value of k is equal to

Options

A. 3
B. 6
C. 9
D. 12

Explanation

To solve the given integral equation and find the value of \( k \), we start with the equation:

\[ \int x^{26} (x-1)^{17} (5x-3) \, dx = \frac{x^{27} (x-1)^{18}}{k} + C \]

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus, we differentiate the left-hand side and the right-hand side:

\[ \frac{d}{dx} \left( \int x^{26} (x-1)^{17} (5x-3) \, dx \right) = x^{26} (x-1)^{17} (5x-3) \]

For the right-hand side, we apply the constant multiple rule and the product rule:

\[ \frac{d}{dx} \left( \frac{x^{27} (x-1)^{18}}{k} \right) = \frac{1}{k} \frac{d}{dx} \left( x^{27} (x-1)^{18} \right) \]

### Step 2: Differentiate \( x^{27} (x-1)^{18} \) using the product rule

Let \( u = x^{27} \) and \( v = (x-1)^{18} \). Then:

\[ \frac{du}{dx} = 27x^{26}, \quad \frac{dv}{dx} = 18(x-1)^{17} \]

Applying the product rule:

\[ \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the derivatives:

\[ \frac{d}{dx} (x^{27} (x-1)^{18}) = x^{27} \cdot 18(x-1)^{17} + (x-1)^{18} \cdot 27x^{26} \]

### Step 3: Combine the derivatives

Now we have:

\[ x^{26} (x-1)^{17} (5x-3) = \frac{1}{k} \left( x^{27} \cdot 18(x-1)^{17} + (x-1)^{18} \cdot 27x^{26} \right) \]

### Step 4: Factor out common terms

We can factor out \( x^{26} (x-1)^{17} \) from both sides:

\[ x^{26} (x-1)^{17} (5x-3) = \frac{1}{k} x^{26} (x-1)^{17} \left( 18x + 27(x-1) \right) \]

### Step 5: Simplify the equation

Cancel \( x^{26} (x-1)^{17} \) from both sides (assuming \( x \neq 0 \) and \( x \neq 1 \)):

\[ 5x - 3 = \frac{1}{k} (18x + 27(x-1)) \]

Expanding the right-hand side:

\[ 5x - 3 = \frac{1}{k} (18x + 27x - 27) = \frac{1}{k} (45x - 27) \]

### Step 6: Cross-multiply to solve for \( k \)

Cross-multiplying gives:

\[ k(5x - 3) = 45x - 27 \]

### Step 7: Equate coefficients

To find \( k \), we can equate coefficients of \( x \) and the constant terms.

From the coefficients of \( x \):

\[ 5k = 45 \implies k = \frac{45}{5} = 9 \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{9} \]