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: Observe the two inverse tangent terms.
The given integrand is \[ \tan^{-1}\left(\frac{x}{x^2+1}\right)+\tan^{-1}\left(\frac{x^2+1}{x}\right) \] Let \[ a=\frac{x}{x^2+1} \] Then, \[ \frac{x^2+1}{x}=\frac{1}{a} \] So the integrand becomes \[ \tan^{-1}a+\tan^{-1}\frac{1}{a} \]

Step 2: Apply inverse tangent identity.
For \(a>0\), \[ \tan^{-1}a+\tan^{-1}\frac{1}{a}=\frac{\pi}{2} \] For \(a<0\), \[ \tan^{-1}a+\tan^{-1}\frac{1}{a}=-\frac{\pi}{2} \] Here, \[ a=\frac{x}{x^2+1} \] Since \[ x^2+1>0 \] for all real \(x\), the sign of \(a\) depends only on \(x\).
Thus, \[ a<0 \quad \text{when} \quad -1\leq x<0 \] and \[ a>0 \quad \text{when} \quad 0<x\leq 2 \]

Step 3: Split the integral at \(x=0\).
Therefore, \[ \int_{-1}^{2}\left[\tan^{-1}\left(\frac{x}{x^2+1}\right)+\tan^{-1}\left(\frac{x^2+1}{x}\right)\right]dx \] \[ = \int_{-1}^{0}\left(-\frac{\pi}{2}\right)dx + \int_{0}^{2}\left(\frac{\pi}{2}\right)dx \]

Step 4: Evaluate the integrals.
\[ \int_{-1}^{0}\left(-\frac{\pi}{2}\right)dx = -\frac{\pi}{2}(0-(-1)) \] \[ =-\frac{\pi}{2} \] Also, \[ \int_{0}^{2}\left(\frac{\pi}{2}\right)dx = \frac{\pi}{2}(2-0) \] \[ =\pi \] Hence, \[ I=-\frac{\pi}{2}+\pi \] \[ I=\frac{\pi}{2} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{\pi}{2}} \]