Question

Evaluate the definite integral $$\int_{1}^{3} \left[ \tan^{-1}\left(\frac{x}{x^2-1}\right) + \tan^{-1}\left(\frac{x^2-1}{x}\right) \right] dx$$

• A. $\pi$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $2\pi$

Hint:

Whenever you see two complicated inverse trigonometric terms added together where the numerator and denominator are flipped, don't try to integrate them individually! They will almost always collapse directly into the constant $\frac{\pi}{2}$ via standard identities.

Answer 1

The Correct Option is A

Step 1: Look at the structure of the integrand.
The two arctangent arguments, $\dfrac{x}{x^2-1}$ and $\dfrac{x^2-1}{x}$, are reciprocals of each other.
Step 2: Recall the complementary identity.
For a positive quantity $u$, $\tan^{-1}u + \tan^{-1}\dfrac{1}{u} = \dfrac{\pi}{2}$.
Step 3: Check the sign on the interval.
For $x$ in $[1,3]$, $x^2 - 1 \ge 0$ and $x > 0$, so both arguments are non-negative and the identity applies.
Step 4: Collapse the integrand to a constant.
The bracket becomes $\tan^{-1}u + \tan^{-1}\dfrac{1}{u} = \dfrac{\pi}{2}$ for every $x$ in the range.
Step 5: Integrate the constant.
$\displaystyle\int_1^3 \dfrac{\pi}{2}\,dx = \dfrac{\pi}{2}\,[x]_1^3 = \dfrac{\pi}{2}(3-1)$.
Step 6: Finish the arithmetic.
$\dfrac{\pi}{2}\times 2 = \pi$.
\[ \boxed{\pi} \]