If `int(1)/((x+1)sqrt(x-1))dx=k.tan^(-1)sqrt((x-1)/(2))+c`
then `k = `

Options

A. `(1)/(sqrt2)`
B. `(1)/(2)`
C. `2sqrt2`
D. `sqrt2`

Answer

To solve the integral \( \int \frac{1}{(x+1)\sqrt{x-1}} \, dx \) and find the value of \( k \) in the equation

\[ \int \frac{1}{(x+1)\sqrt{x-1}} \, dx = k \tan^{-1} \left( \frac{\sqrt{x-1}}{2} \right) + c, \]

we will follow these steps:

### Step 1: Substitution
Let \( y^2 = x - 1 \). Then, we have:
\[ x = y^2 + 1. \]
Differentiating both sides gives:
\[ dx = 2y \, dy. \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ \int \frac{1}{(y^2 + 1 + 1)\sqrt{y^2}} \cdot 2y \, dy = \int \frac{2y}{(y^2 + 2)y} \, dy. \]
The \( y \) in the numerator and denominator cancels out:
\[ \int \frac{2}{y^2 + 2} \, dy. \]

### Step 3: Simplify the Integral
The integral simplifies to:
\[ 2 \int \frac{1}{y^2 + 2} \, dy. \]
This can be rewritten as:
\[ 2 \int \frac{1}{2} \cdot \frac{1}{\frac{y^2}{2} + 1} \, dy = \int \frac{1}{\frac{y^2}{2} + 1} \, dy. \]

### Step 4: Use the Arctangent Formula
The integral \( \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \) applies here with \( a^2 = 2 \) (thus \( a = \sqrt{2} \)):
\[ = \sqrt{2} \tan^{-1} \left( \frac{y}{\sqrt{2}} \right) + C. \]

### Step 5: Substitute Back
Substituting back \( y = \sqrt{x - 1} \):
\[ = \sqrt{2} \tan^{-1} \left( \frac{\sqrt{x - 1}}{\sqrt{2}} \right) + C. \]

### Step 6: Compare with Given Equation
Now, we compare this with the given equation:
\[ k \tan^{-1} \left( \frac{\sqrt{x - 1}}{2} \right). \]
To match the forms, we can rewrite \( \tan^{-1} \left( \frac{\sqrt{x - 1}}{\sqrt{2}} \right) \) as:
\[ \sqrt{2} \tan^{-1} \left( \frac{\sqrt{x - 1}}{2} \right). \]
Thus, we find that:
\[ k = \sqrt{2}. \]

### Final Answer
Therefore, the value of \( k \) is:
\[ \boxed{\sqrt{2}}. \]