When x 0,then int cos^{-1}left(frac{1-x^{2} | vedclass

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(D) Given,for $x > 0$,we need to evaluate $I = \int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}ight) dx$. Using the trigonometric substitution $x = 	an

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$2x 	an^{-1} x - \log(1+x^{2}) + C$

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(D) Given,for $x > 0$,we need to evaluate $I = \int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}ight) dx$. Using the trigonometric substitution $x = 	an 	heta$,we know that for $x > 0$,$\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}ight) = 2 	an^{-1} x$. Substituting this into the integral,we get $I = \int 2 	an^{-1} x dx = 2 \int 	an^{-1} x dx$. Using integration by parts,let $u = 	an^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^{2}} dx$ and $v = x$. $I = 2 \left( x 	an^{-1} x - \int \frac{x}{1+x^{2}} dx ight)$. To solve $\int \frac{x}{1+x^{2}} dx$,let $t = 1+x^{2}$,then $dt = 2x dx$,so $x dx = \frac{1}{2} dt$. Thus,$\int \frac{x}{1+x^{2}} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log(1+x^{2})$. Substituting this back,$I = 2x 	an^{-1} x - 2 \left( \frac{1}{2} \log(1+x^{2}) ight) + C = 2x 	an^{-1} x - \log(1+x^{2}) + C$.

When $x > 0$,then $\int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) dx$ is

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