Number of values of `x` satisfying the equation `int_(-1)^x (8t^2+28/3t+4) \ dt=((3/2)x+1)/(log_(x+1)sqrt(x+1)),` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation

\[ \int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) \, dt = \frac{\frac{3}{2}x + 1}{\log_{(x+1)}\sqrt{x+1}}, \]

we will follow these steps:

### Step 1: Evaluate the Left-Hand Side (LHS)

We need to compute the integral on the left-hand side:

\[ \int (8t^2 + \frac{28}{3}t + 4) \, dt. \]

Calculating the integral:

\[ \int (8t^2) \, dt = \frac{8}{3}t^3, \]
\[ \int \left(\frac{28}{3}t\right) \, dt = \frac{28}{6}t^2 = \frac{14}{3}t^2, \]
\[ \int 4 \, dt = 4t. \]

Thus, the indefinite integral is:

\[ \frac{8}{3}t^3 + \frac{14}{3}t^2 + 4t + C. \]

Now, we evaluate the definite integral from \(-1\) to \(x\):

\[ \int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) \, dt = \left[\frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x\right] - \left[\frac{8}{3}(-1)^3 + \frac{14}{3}(-1)^2 + 4(-1)\right]. \]

Calculating the value at \(t = -1\):

\[ \frac{8}{3}(-1)^3 + \frac{14}{3}(-1)^2 + 4(-1) = \frac{-8}{3} + \frac{14}{3} - 4 = \frac{-8 + 14 - 12}{3} = \frac{-6}{3} = -2. \]

Thus, the LHS becomes:

\[ \frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2. \]

### Step 2: Simplify the Right-Hand Side (RHS)

The right-hand side is:

\[ \frac{\frac{3}{2}x + 1}{\log_{(x+1)}\sqrt{x+1}}. \]

Using the property of logarithms, we can rewrite \(\log_{(x+1)}\sqrt{x+1}\) as:

\[ \log_{(x+1)}(x+1)^{1/2} = \frac{1}{2}\log_{(x+1)}(x+1) = \frac{1}{2}. \]

Thus, the RHS simplifies to:

\[ \frac{\frac{3}{2}x + 1}{\frac{1}{2}} = 3x + 2. \]

### Step 3: Set the LHS equal to the RHS

Now we have the equation:

\[ \frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2 = 3x + 2. \]

Subtract \(3x + 2\) from both sides:

\[ \frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2 - 3x - 2 = 0. \]

This simplifies to:

\[ \frac{8}{3}x^3 + \frac{14}{3}x^2 + (4 - 3)x = 0, \]

or

\[ \frac{8}{3}x^3 + \frac{14}{3}x^2 + x = 0. \]

### Step 4: Factor the equation

Factoring out \(x\):

\[ x\left(\frac{8}{3}x^2 + \frac{14}{3}x + 1\right) = 0. \]

This gives us one solution:

\[ x = 0. \]

Now we need to find the roots of the quadratic:

\[ \frac{8}{3}x^2 + \frac{14}{3}x + 1 = 0. \]

### Step 5: Use the quadratic formula

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = \frac{8}{3}\), \(b = \frac{14}{3}\), and \(c = 1\):

\[ b^2 - 4ac = \left(\frac{14}{3}\right)^2 - 4 \cdot \frac{8}{3} \cdot 1 = \frac{196}{9} - \frac{32}{3} = \frac{196}{9} - \frac{96}{9} = \frac{100}{9}. \]

Now substituting back into the formula:

\[ x = \frac{-\frac{14}{3} \pm \sqrt{\frac{100}{9}}}{2 \cdot \frac{8}{3}} = \frac{-\frac{14}{3} \pm \frac{10}{3}}{\frac{16}{3}} = \frac{-14 \pm 10}{16}. \]

Calculating the two possible values:

1. \(x = \frac{-4}{16} = -\frac{1}{4}\)
2. \(x = \frac{-24}{16} = -\frac{3}{2}\)

### Step 6: Check the validity of solutions

We have three potential solutions: \(x = 0\), \(x = -\frac{1}{4}\), and \(x = -\frac{3}{2}\).

However, we need to ensure that \(x + 1 > 0\) for the logarithm to be defined. Thus, \(x > -1\).

Only \(x = 0\) and \(x = -\frac{1}{4}\) satisfy this condition.

### Conclusion

The number of values of \(x\) satisfying the original equation is **2**.

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