For \( x>0 \), if \( \int \frac{1}{x^{2}+5x+7} \, dx = \frac{2}{\sqrt{3}}F(x)+k \) and \( F\left(-\frac{5}{2}\right)=0 \), then \( \sin(F(x)) = \)
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Once you know that \( \tan(F(x)) = \frac{2x+5}{\sqrt{3}} \), the hypotenuse must contain the square root of the original quadratic denominator expression. This lets you identify the correct option immediately by looking at the denominators.
Step 1: Complete the square in the denominator. \[ x^2+5x+7=\left(x+\tfrac{5}{2}\right)^2+7-\tfrac{25}{4}=\left(x+\tfrac{5}{2}\right)^2+\left(\tfrac{\sqrt{3}}{2}\right)^2 \] Step 2: Apply the standard formula. Using $\int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\frac{u}{a}$ with $u=x+\frac{5}{2}$ and $a=\frac{\sqrt{3}}{2}$: \[ I=\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{2x+5}{\sqrt{3}}\right)+c \] Step 3: Identify $F(x)$. Comparing with $\frac{2}{\sqrt{3}}F(x)+k$ gives $F(x)=\tan^{-1}\!\left(\frac{2x+5}{\sqrt{3}}\right)$. The check $F(-\tfrac{5}{2})=\tan^{-1}0=0$ matches. Step 4: Set up a right triangle. Let $\theta=F(x)$ so $\tan\theta=\frac{2x+5}{\sqrt{3}}$. Take opposite $=2x+5$ and adjacent $=\sqrt{3}$. Step 5: Find the hypotenuse. \[ \sqrt{(2x+5)^2+3}=\sqrt{4x^2+20x+28}=2\sqrt{x^2+5x+7} \] Step 6: Read off the sine. \[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2x+5}{2\sqrt{x^2+5x+7}} \] \[ \boxed{\dfrac{2x+5}{2\sqrt{x^2+5x+7}}} \]