The value of the integral \int\limits_1/2^2 \tan ^ - 1x \over xdx is equal to :

Definite Integration PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q36 with Solution

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Correct answer: A

Let $I=\int_{1/2}^{2}\frac{\tan^{-1}x}{x}\,dx.$ We use the standard identity $\tan^{-1}x+\tan^{-1}\left(\frac1x\right)=\frac\pi2 \quad (x>0).$ Since the interval $\left[\frac12,2\right]$ lies in $x>0$ , this identity is valid throughout.
Make the substitution $x=\frac1t\quad \Rightarrow \quad dx=-\frac1{t^2}dt.$ Then

=\int_{2}^{1/2}\frac{\tan^{-1}(1/t)}{1/t}\left(-\frac1{t^2}\right)dt =\int_{1/2}^{2}\frac{\tan^{-1}(1/t)}{t}\,dt.$$ Renaming $t$ as $x$, $$I=\int_{1/2}^{2}\frac{\tan^{-1}(1/x)}{x}\,dx.$$ 3. Add the two expressions for $I$: $$2I=\int_{1/2}^{2}\frac{\tan^{-1}x+\tan^{-1}(1/x)}{x}\,dx.$$ Using $$\tan^{-1}x+\tan^{-1}(1/x)=\frac\pi2,$$ we get $$2I=\int_{1/2}^{2}\frac{\pi/2}{x}\,dx =\frac\pi2\int_{1/2}^{2}\frac{dx}{x}.$$ 4. Evaluate the logarithmic integral: $$\int_{1/2}^{2}\frac{dx}{x}=\ln 2-\ln\left(\frac12\right)=\ln 2+\ln 2=2\ln 2.$$ So, $$2I=\frac\pi2\cdot 2\ln 2=\pi\ln 2.$$ Hence, $$I=\frac\pi2\ln 2.$$ 5. Compare with the options: - A: $\dfrac\pi2\ln 2$ ✅ - B: $\dfrac\pi4\ln 2$ - C: $\dfrac12\ln 2$ - D: $\pi\ln 2$ Therefore, the correct option is **A**.

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Correct answer: A

Let $I=\int_{1/2}^{2}\frac{\tan^{-1}x}{x}\,dx.$ We use the standard identity $\tan^{-1}x+\tan^{-1}\left(\frac1x\right)=\frac\pi2 \quad (x>0).$ Since the interval $\left[\frac12,2\right]$ lies in $x>0$ , this identity is valid throughout.
Make the substitution $x=\frac1t\quad \Rightarrow \quad dx=-\frac1{t^2}dt.$ Then

=\int_{2}^{1/2}\frac{\tan^{-1}(1/t)}{1/t}\left(-\frac1{t^2}\right)dt =\int_{1/2}^{2}\frac{\tan^{-1}(1/t)}{t}\,dt.$$ Renaming $t$ as $x$, $$I=\int_{1/2}^{2}\frac{\tan^{-1}(1/x)}{x}\,dx.$$ 3. Add the two expressions for $I$: $$2I=\int_{1/2}^{2}\frac{\tan^{-1}x+\tan^{-1}(1/x)}{x}\,dx.$$ Using $$\tan^{-1}x+\tan^{-1}(1/x)=\frac\pi2,$$ we get $$2I=\int_{1/2}^{2}\frac{\pi/2}{x}\,dx =\frac\pi2\int_{1/2}^{2}\frac{dx}{x}.$$ 4. Evaluate the logarithmic integral: $$\int_{1/2}^{2}\frac{dx}{x}=\ln 2-\ln\left(\frac12\right)=\ln 2+\ln 2=2\ln 2.$$ So, $$2I=\frac\pi2\cdot 2\ln 2=\pi\ln 2.$$ Hence, $$I=\frac\pi2\ln 2.$$ 5. Compare with the options: - A: $\dfrac\pi2\ln 2$ ✅ - B: $\dfrac\pi4\ln 2$ - C: $\dfrac12\ln 2$ - D: $\pi\ln 2$ Therefore, the correct option is **A**.