If `(x+1)^(2)/(x^(3)+x)=A/x+(Bx+C)/(x^(2)+1) " then "Sin^(-1)(A/C)=`

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem, we need to express the given fraction \(\frac{(x+1)^{2}}{x^{3}+x}\) in terms of partial fractions. Let's go through the solution step by step.

### Step 1: Factor the Denominator
First, we need to factor the denominator \(x^3 + x\):
\[ x^3 + x = x(x^2 + 1) \]
So, we can rewrite the expression as:
\[ \frac{(x+1)^{2}}{x(x^{2}+1)} \]

### Step 2: Set Up the Partial Fraction Decomposition
We want to express this as:
\[ \frac{(x+1)^{2}}{x(x^{2}+1)} = \frac{A}{x} + \frac{Bx + C}{x^{2}+1} \]
where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 3: Combine the Right Side
To combine the right side over a common denominator:
\[ \frac{A}{x} + \frac{Bx + C}{x^{2}+1} = \frac{A(x^{2}+1) + (Bx+C)x}{x(x^{2}+1)} \]
This simplifies to:
\[ \frac{Ax^{2} + A + Bx^{2} + Cx}{x(x^{2}+1)} = \frac{(A + B)x^{2} + Cx + A}{x(x^{2}+1)} \]

### Step 4: Set the Numerators Equal
Now we equate the numerators:
\[ (A + B)x^{2} + Cx + A = (x + 1)^{2} \]
Expanding \((x + 1)^{2}\):
\[ (x + 1)^{2} = x^{2} + 2x + 1 \]
Thus, we have:
\[ (A + B)x^{2} + Cx + A = x^{2} + 2x + 1 \]

### Step 5: Compare Coefficients
Now we can compare the coefficients from both sides:
1. For \(x^{2}\): \(A + B = 1\)
2. For \(x\): \(C = 2\)
3. For the constant term: \(A = 1\)

### Step 6: Solve the Equations
From \(A = 1\):
\[ A = 1 \]
Substituting \(A\) into the first equation:
\[ 1 + B = 1 \implies B = 0 \]
And from the second equation:
\[ C = 2 \]

### Step 7: Find \(A\) and \(C\)
We have:
\[ A = 1, \quad C = 2 \]

### Step 8: Calculate \(\sin^{-1}\left(\frac{A}{C}\right)\)
Now we need to find:
\[ \sin^{-1}\left(\frac{A}{C}\right) = \sin^{-1}\left(\frac{1}{2}\right) \]
The value of \(\sin^{-1}\left(\frac{1}{2}\right)\) is:
\[ \frac{\pi}{6} \]

### Final Answer
Thus, the final answer is:
\[ \sin^{-1}\left(\frac{A}{C}\right) = \frac{\pi}{6} \]

---