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(B) First,evaluate the integral on the $LHS$: $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = [\frac{8}{3}t^3 + \frac{14}{3}t^2 + 4t]_{-1}^{x} = (\frac

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$1$

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(B) First,evaluate the integral on the $LHS$: $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = [\frac{8}{3}t^3 + \frac{14}{3}t^2 + 4t]_{-1}^{x} = (\frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x) - (-\frac{8}{3} + \frac{14}{3} - 4) = \frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2$. For the $RHS$,note that $\log_{(x+1)} \sqrt{x+1} = \log_{(x+1)} (x+1)^{1/2} = \frac{1}{2}$. Thus,the equation becomes $\frac{8}{3}x^3 + \frac{14}{3}x^2 + 4x + 2 = \frac{\frac{3}{2}x + 1}{1/2} = 3x + 2$. Subtracting $3x + 2$ from both sides: $\frac{8}{3}x^3 + \frac{14}{3}x^2 + x = 0$. Multiplying by $3$: $8x^3 + 14x^2 + 3x = 0$. Factoring: $x(8x^2 + 14x + 3) = 0$,which gives $x(4x + 1)(2x + 3) = 0$. The roots are $x = 0$,$x = -1/4$,and $x = -3/2$. However,the logarithm base $x+1$ must satisfy $x+1 > 0$ and $x+1 
eq 1$,so $x > -1$ and $x 
eq 0$. Checking the roots: $x=0$ is excluded by the base condition. $x=-3/2$ is excluded as $-3/2 Only $x = -1/4$ satisfies the condition $x > -1$ and $x 
eq 0$. Thus,there is only $1$ value of $x$.

Number of values of $x$ satisfying the equation $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = \frac{(\frac{3}{2})x + 1}{\log_{(x+1)} \sqrt{x+1}}$.

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