int frac{tan^{-1} x - cot^{-1} x}{tan^{-1} x | vedclass

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(D) We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.Also,$\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$.Substituting

Vedclass · Accepted Answer

$\frac{4}{\pi} x 	an^{-1} x - \frac{2}{\pi} \ln(1 + x^2) - x + c$

Vedclass · Answer

(D) We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.Also,$\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$.Substituting these into the integral:$I = \int \frac{	an^{-1} x - (\frac{\pi}{2} - 	an^{-1} x)}{\pi/2} \, dx$$I = \frac{2}{\pi} \int (2 	an^{-1} x - \frac{\pi}{2}) \, dx$$I = \frac{4}{\pi} \int 	an^{-1} x \, dx - \int 1 \, dx$Using integration by parts for $\int 	an^{-1} x \, dx$ (taking $u = 	an^{-1} x$ and $dv = dx$):$\int 	an^{-1} x \, dx = x 	an^{-1} x - \int \frac{x}{1 + x^2} \, dx = x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)$.Substituting this back:$I = \frac{4}{\pi} [x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)] - x + c$$I = \frac{4}{\pi} x 	an^{-1} x - \frac{2}{\pi} \ln(1 + x^2) - x + c$.

$\int \frac{\tan^{-1} x - \cot^{-1} x}{\tan^{-1} x + \cot^{-1} x} \, dx$ is equal to:

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