The value of int {frac{{sqrt {{x^2} - {a^2}} | vedclass

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Question

(A) Let $I = \int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx}$.Substitute $\sqrt {{x^2} - {a^2}} = t$,then ${x^2} - {a^2} = {t^2}$,which implies ${x^2} = {

Vedclass · Accepted Answer

$\sqrt {{x^2} - {a^2}} - a{	an ^{ - 1}}\left( {\frac{{\sqrt {{x^2} - {a^2}} }}{a}} ight) + C$

Vedclass · Answer

(A) Let $I = \int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx}$.Substitute $\sqrt {{x^2} - {a^2}} = t$,then ${x^2} - {a^2} = {t^2}$,which implies ${x^2} = {a^2} + {t^2}$.Differentiating both sides,$2xdx = 2tdt$,so $xdx = tdt$.Now,multiply the numerator and denominator by $x$:$I = \int {\frac{{\sqrt {{x^2} - {a^2}} \cdot x}}{{{x^2}}}dx} = \int {\frac{t}{{{a^2} + {t^2}}} \cdot tdt} = \int {\frac{{{t^2}}}{{{a^2} + {t^2}}}dt}$.$I = \int {\left( {\frac{{{t^2} + {a^2} - {a^2}}}{{{a^2} + {t^2}}}} ight)dt} = \int {\left( {1 - \frac{{{a^2}}}{{{a^2} + {t^2}}}} ight)dt}$.$I = \int {1dt} - {a^2}\int {\frac{1}{{{a^2} + {t^2}}}dt} = t - {a^2} \cdot \frac{1}{a}{	an ^{ - 1}}\left( {\frac{t}{a}} ight) + C$.$I = t - a{	an ^{ - 1}}\left( {\frac{t}{a}} ight) + C$.Substituting back $t = \sqrt {{x^2} - {a^2}}$,we get:$I = \sqrt {{x^2} - {a^2}} - a{	an ^{ - 1}}\left( {\frac{{\sqrt {{x^2} - {a^2}} }}{a}} ight) + C$.

The value of $\int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx} $ is:

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