Let F(x) be the primitive of `(3x+2)/sqrt(x-9)`w.r.t.x. If `F(10)=60` then the value of `F(13)`

Answer

To solve the problem step by step, we need to find the integral of the function \( F(x) = \int \frac{3x + 2}{\sqrt{x - 9}} \, dx \) and then use the given condition \( F(10) = 60 \) to find the constant of integration. Finally, we will evaluate \( F(13) \).

### Step 1: Rewrite the Integral
We start with the integral:
\[ F(x) = \int \frac{3x + 2}{\sqrt{x - 9}} \, dx \]

### Step 2: Substitution
Let \( t = \sqrt{x - 9} \). Then, we have:
\[ x - 9 = t^2 \quad \Rightarrow \quad x = t^2 + 9 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 3: Substitute in the Integral
Now, substitute \( x \) and \( dx \) into the integral:
\[ F(x) = \int \frac{3(t^2 + 9) + 2}{t} \cdot 2t \, dt \]
This simplifies to:
\[ F(x) = \int (3(t^2 + 9) + 2) \cdot 2 \, dt = \int (6t^2 + 60 + 4) \, dt = \int (6t^2 + 64) \, dt \]

### Step 4: Integrate
Now, we can integrate:
\[ F(x) = 6 \cdot \frac{t^3}{3} + 64t + C = 2t^3 + 64t + C \]

### Step 5: Substitute Back for \( t \)
Substituting back \( t = \sqrt{x - 9} \):
\[ F(x) = 2(\sqrt{x - 9})^3 + 64\sqrt{x - 9} + C = 2(x - 9)^{3/2} + 64(x - 9)^{1/2} + C \]

### Step 6: Use the Condition \( F(10) = 60 \)
Now we use the condition \( F(10) = 60 \):
\[ F(10) = 2(10 - 9)^{3/2} + 64(10 - 9)^{1/2} + C = 2(1) + 64(1) + C = 2 + 64 + C = 66 + C \]
Setting this equal to 60 gives:
\[ 66 + C = 60 \quad \Rightarrow \quad C = 60 - 66 = -6 \]

### Step 7: Write the Final Expression for \( F(x) \)
Thus, we have:
\[ F(x) = 2(x - 9)^{3/2} + 64(x - 9)^{1/2} - 6 \]

### Step 8: Find \( F(13) \)
Now we can find \( F(13) \):
\[ F(13) = 2(13 - 9)^{3/2} + 64(13 - 9)^{1/2} - 6 \]
Calculating:
\[ F(13) = 2(4)^{3/2} + 64(4)^{1/2} - 6 = 2 \cdot 8 + 64 \cdot 2 - 6 = 16 + 128 - 6 = 138 \]

### Final Answer
Thus, the value of \( F(13) \) is:
\[ \boxed{138} \]