If `int(dx)/(xsqrt(5x^(2)-3))=Ktan^(-1)f(x)+C` then

Options

A. `f(x) = sqrt((5)/(3)x^(2)-1),K=(1)/(sqrt(5))`
B. `f(x) = sqrt((5)/(3)x^(2)-1),K=(1)/(sqrt(3))`
C. `f(x) = (1)/(2)sqrt(5x^(2)-3,)K=(1)/(sqrt(5)`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{dx}{x \sqrt{5x^2 - 3}}, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ I = \int \frac{1}{x \sqrt{5x^2 - 3}} \, dx. \]

### Step 2: Substitution
Let \( t^2 = 5x^2 - 3 \). Then, differentiating both sides gives:

\[ 2t \, dt = 10x \, dx \implies dx = \frac{t \, dt}{5x}. \]

Now, we need to express \( x \) in terms of \( t \):

\[ x^2 = \frac{t^2 + 3}{5} \implies x = \sqrt{\frac{t^2 + 3}{5}}. \]

### Step 3: Substitute in the Integral
Substituting \( dx \) and \( x \) into the integral:

\[ I = \int \frac{1}{\sqrt{\frac{t^2 + 3}{5}} \sqrt{t^2}} \cdot \frac{t \, dt}{5 \sqrt{\frac{t^2 + 3}{5}}} = \int \frac{t \, dt}{5 \cdot \frac{t^2 + 3}{5} \cdot t} = \int \frac{dt}{t^2 + 3}. \]

### Step 4: Integrate
The integral

\[ \int \frac{dt}{t^2 + 3} \]

can be solved using the formula

\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C. \]

Here, \( a^2 = 3 \) so \( a = \sqrt{3} \):

\[ I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + C. \]

### Step 5: Substitute Back
Now substituting back \( t = \sqrt{5x^2 - 3} \):

\[ I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{5x^2 - 3}}{\sqrt{3}} \right) + C. \]

### Step 6: Compare with Given Form
We are given that

\[ I = K \tan^{-1}(f(x)) + C. \]

From our result, we can identify:

\[ K = \frac{1}{\sqrt{3}}, \quad f(x) = \frac{\sqrt{5x^2 - 3}}{\sqrt{3}}. \]

### Final Answer
Thus, we find:

\[ K = \frac{1}{\sqrt{3}}, \quad f(x) = \frac{\sqrt{5x^2 - 3}}{\sqrt{3}}. \]

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