int frac{dx}{(1 + x^2)sqrt{p^2 + q^2(tan^{-1} | vedclass

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Question

(A) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{p^2 + q^2(	an^{-1}x)^2}}$.Substitute $t = q	an^{-1}x$.Then $dt = \frac{q}{1 + x^2} dx$,which implies $\fr

Vedclass · Accepted Answer

$\frac{1}{q}\log |q	an^{-1}x + \sqrt{p^2 + q^2(	an^{-1}x)^2}| + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{p^2 + q^2(	an^{-1}x)^2}}$.Substitute $t = q	an^{-1}x$.Then $dt = \frac{q}{1 + x^2} dx$,which implies $\frac{dx}{1 + x^2} = \frac{dt}{q}$.Substituting these into the integral,we get:$I = \int \frac{1}{\sqrt{p^2 + t^2}} \cdot \frac{dt}{q} = \frac{1}{q} \int \frac{dt}{\sqrt{p^2 + t^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log |x + \sqrt{x^2 + a^2}| + c$,we have:$I = \frac{1}{q} \log |t + \sqrt{t^2 + p^2}| + c$.Substituting $t = q	an^{-1}x$ back into the expression:$I = \frac{1}{q} \log |q	an^{-1}x + \sqrt{q^2(	an^{-1}x)^2 + p^2}| + c$.

$\int \frac{dx}{(1 + x^2)\sqrt{p^2 + q^2(\tan^{-1}x)^2}} = $

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