Question

\( \int_{-1}^{0} \frac{dx}{x^2 + x + 2} \) is equal to:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \pi \)
• D. \( 0 \)
• E. \( -\pi \)

Hint:

For symmetric limits like \( [-a,a] \), always check if substitution simplifies the integral into a standard inverse trigonometric form.

Answer 1

The Correct Option is A

Concept: To evaluate integrals of the form: \[ \int \frac{dx}{x^2 + ax + b} \] we complete the square and use: \[ \int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) \]

Step 1: Complete the square.
\[ x^2 + x + 2 = \left(x + \frac{1}{2}\right)^2 + \frac{7}{4} \] So the integral becomes: \[ \int_{-1}^{0} \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{7}}{2}\right)^2} \]

Step 2: Apply standard formula.
\[ = \left[ \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{2x+1}{\sqrt{7}}\right) \right]_{-1}^{0} \]

Step 3: Substitute limits.
At \( x = 0 \): \[ \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \] At \( x = -1 \): \[ \tan^{-1}\left(\frac{-1}{\sqrt{7}}\right) = -\tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

Step 4: Simplify the expression.
\[ = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) - \left(-\tan^{-1}\left(\frac{1}{\sqrt{7}}\right)\right) \right] \] \[ = \frac{4}{\sqrt{7}} \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

Step 5: Evaluate using standard value.
\[ \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) = \frac{\pi}{6} \] \[ \Rightarrow \frac{4}{\sqrt{7}} \cdot \frac{\pi}{6} \] \[ = \frac{\pi}{2} \]

Step 6: Final answer.
\[ \boxed{\frac{\pi}{2}} \]