Question

The value of the integral \(∫_{\frac 12}^2 \frac{tan^{-1}x}{x}dx\) is equal to

• A. $ \frac{π}2 log_e⁡2$
• B. $ 2 log_e⁡2$
• C. $ \frac{1}2 log_e⁡2$
• D. $ \frac{π}4 log_e⁡2$

Answer 1

The Correct Option is A

(A)The given integral is: \[ I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx. \] (B)Use the substitution \( x = \frac{1}{t} \), so \( dx = -\frac{1}{t^2} dt \). The limits of integration change as follows: \[ x = \frac{1}{2} \implies t = 2, \quad x = 2 \implies t = \frac{1}{2}. \] Substituting, the integral becomes: \[ I = \int_{2}^{1/2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{\frac{1}{t}} \cdot \left(-\frac{1}{t^2}\right) dt. \] (C)Simplify: \[ I = \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{t}\right)}{t} dt. \] (D)Add the original integral and its substitution: \[ 2I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx + \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{x}\right)}{x} \, dx. \] (E)Use the property \( \tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \) for \( x > 0 \). Thus: \[ 2I = \int_{1/2}^2 \frac{\frac{\pi}{2}}{x} \, dx. \] (F) Simplify: \[ 2I = \frac{\pi}{2} \int_{1/2}^2 \frac{1}{x} \, dx. \] The integral of \( \frac{1}{x} \) is \( \log_e x \): \[ 2I = \frac{\pi}{2} \left[\log_e x\right]_{1/2}^2. \] (G) Evaluate: \[ 2I = \frac{\pi}{2} \left(\log_e 2 - \log_e \frac{1}{2}\right). \] Simplify \( \log_e \frac{1}{2} = -\log_e 2 \): \[ 2I = \frac{\pi}{2} \left(\log_e 2 - (-\log_e 2)\right) = \frac{\pi}{2} (2\log_e 2). \] Divide by 2: \[ I = \frac{\pi}{2} \log_e 2. \]