Question:
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 \( \sin(F(x)) = \)
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 \( \sin(F(x)) = \)
Show Hint
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.
Updated On: Jun 7, 2026
- \( \frac{2x-5}{\sqrt{3}} \)
- \( \frac{2x+5}{2\sqrt{x^{2}+5x+7}} \)
- \( \frac{2\sqrt{x^{2}+5x+7}}{2x+5} \)
- \( \frac{2\sqrt{x^{2}+5x+7}}{\sqrt{3}} \)
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The Correct Option is B
Solution and Explanation
Concept:
To integrate a rational expression with a quadratic denominator, we first complete the square in the denominator to transform it into the standard form \( \int \frac{1}{u^2 + a^2} du = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) \).
Step 1: Completing the square in the denominator.
\[ x^2 + 5x + 7 = \left(x + \frac{5}{2}\right)^2 + 7 - \frac{25}{4} = \left(x + \frac{5}{2}\right)^2 + \frac{3}{4} = \left(x + \frac{5}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]
Step 2: Performing the integration.
Using the standard formula with \( u = x + \frac{5}{2} \) and \( a = \frac{\sqrt{3}}{2} \): \[ I = \frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\frac{x + \frac{5}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+5}{\sqrt{3}}\right) \]
Step 3: Identifying function \( F(x) \).
Comparing this with the given form \( \frac{2}{\sqrt{3}}F(x) + k \), we find: \[ F(x) = \tan^{-1}\left(\frac{2x+5}{\sqrt{3}}\right) \] Let us check the condition: \( F(-5/2) = \tan^{-1}(0) = 0 \), which matches perfectly.
Step 4: Calculating \( \sin(F(x)) \).
Let \( \theta = F(x) \implies \tan\theta = \frac{2x+5}{\sqrt{3}} \). Constructing a right-angled triangle:
• Opposite side \( = 2x + 5 \)
• Adjacent side \( = \sqrt{3} \)
• Hypotenuse \( = \sqrt{(2x+5)^2 + (\sqrt{3})^2} = \sqrt{4x^2 + 20x + 25 + 3} = \sqrt{4(x^2+5x+7)} = 2\sqrt{x^2+5x+7} \)
Therefore, the sine value is: \[ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2x+5}{2\sqrt{x^2+5x+7}} \] This matches option (B) perfectly.
Step 1: Completing the square in the denominator.
\[ x^2 + 5x + 7 = \left(x + \frac{5}{2}\right)^2 + 7 - \frac{25}{4} = \left(x + \frac{5}{2}\right)^2 + \frac{3}{4} = \left(x + \frac{5}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]
Step 2: Performing the integration.
Using the standard formula with \( u = x + \frac{5}{2} \) and \( a = \frac{\sqrt{3}}{2} \): \[ I = \frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\frac{x + \frac{5}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+5}{\sqrt{3}}\right) \]
Step 3: Identifying function \( F(x) \).
Comparing this with the given form \( \frac{2}{\sqrt{3}}F(x) + k \), we find: \[ F(x) = \tan^{-1}\left(\frac{2x+5}{\sqrt{3}}\right) \] Let us check the condition: \( F(-5/2) = \tan^{-1}(0) = 0 \), which matches perfectly.
Step 4: Calculating \( \sin(F(x)) \).
Let \( \theta = F(x) \implies \tan\theta = \frac{2x+5}{\sqrt{3}} \). Constructing a right-angled triangle:
• Opposite side \( = 2x + 5 \)
• Adjacent side \( = \sqrt{3} \)
• Hypotenuse \( = \sqrt{(2x+5)^2 + (\sqrt{3})^2} = \sqrt{4x^2 + 20x + 25 + 3} = \sqrt{4(x^2+5x+7)} = 2\sqrt{x^2+5x+7} \)
Therefore, the sine value is: \[ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2x+5}{2\sqrt{x^2+5x+7}} \] This matches option (B) perfectly.
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