`int (dx)/((1+sqrtx)^(8))=-(1)/(3(1+sqrtx)^(k_(1)))+ (2)/(7(1+sqrtx)^(k_(2)))+C,` then:

Options

A. `K_(1)=5`
B. `k _(1)=6`
C. `k _(2)=7`
D. `k _(2)=8`

Answer

To solve the integral

\[ I = \int \frac{dx}{(1 + \sqrt{x})^8} \]

we will use substitution and integration techniques.

### Step 1: Substitution
Let \( t = 1 + \sqrt{x} \). Then, we differentiate to find \( dx \):
\[ \sqrt{x} = t - 1 \implies x = (t - 1)^2 \]
Differentiating \( x \) with respect to \( t \):
\[ dx = 2(t - 1) dt \]

### Step 2: Rewrite the Integral
Now, substituting \( dx \) and \( \sqrt{x} \) into the integral:
\[ I = \int \frac{2(t - 1) dt}{t^8} \]
This can be simplified as:
\[ I = 2 \int \left( \frac{t}{t^8} - \frac{1}{t^8} \right) dt = 2 \int \left( t^{-7} - t^{-8} \right) dt \]

### Step 3: Integrate
Now we integrate term by term:
\[ I = 2 \left( \int t^{-7} dt - \int t^{-8} dt \right) \]
Using the formula \( \int t^n dt = \frac{t^{n+1}}{n+1} + C \):
\[ \int t^{-7} dt = \frac{t^{-6}}{-6} = -\frac{1}{6} t^{-6} \]
\[ \int t^{-8} dt = \frac{t^{-7}}{-7} = -\frac{1}{7} t^{-7} \]
Thus, we have:
\[ I = 2 \left( -\frac{1}{6} t^{-6} + \frac{1}{7} t^{-7} \right) + C \]
\[ I = -\frac{1}{3} t^{-6} + \frac{2}{7} t^{-7} + C \]

### Step 4: Substitute Back
Now substituting back \( t = 1 + \sqrt{x} \):
\[ I = -\frac{1}{3 (1 + \sqrt{x})^6} + \frac{2}{7 (1 + \sqrt{x})^7} + C \]

### Step 5: Identify \( k_1 \) and \( k_2 \)
From the final expression, we can see that:
\[ k_1 = 6 \quad \text{and} \quad k_2 = 7 \]

Thus, the values of \( k_1 \) and \( k_2 \) are:
\[ k_1 = 6, \quad k_2 = 7 \]