Question

\(\displaystyle \int_{-1}^{2}\left[\tan^{-1}\left(\frac{x}{x^2+1}\right)+\tan^{-1}\left(\frac{x^2+1}{x}\right)\right]dx=\)

• A. \(\dfrac{\pi}{4}\)
• B. \(\dfrac{3\pi}{4}\)
• C. \(\pi/4\)
• D. \(\dfrac{\pi}{2}\)

Hint:

For inverse tangent expressions, remember: \[ \tan^{-1}a+\tan^{-1}\frac{1}{a} = \begin{cases} \dfrac{\pi}{2}, & a>0 -\dfrac{\pi}{2}, & a<0 \end{cases} \] So always check the sign of \(a\) before applying the identity.

Answer 1

The Correct Option is D

Step 1: Recognize the complementary angle relationship.
Let θ = tan⁻¹(x/(x²+1)). Then tan θ = x/(x²+1). The other term is tan⁻¹((x²+1)/x) = tan⁻¹(1/tan θ) = π/2 – θ when tan θ>0, and –π/2 – θ when tan θ<0.

Step 2: Sum simplifies directly.
The sum = π/2 for x>0, and –π/2 for x<0.

Step 3: Integrate over the interval [–1, 2].
∫₋₁² = ∫₋₁⁰ (–π/2)dx + ∫₀² (π/2)dx = (–π/2)(0–(–1)) + (π/2)(2–0) = –π/2 + π = π/2.

Step 4: Final Answer:
π/2.