int left(frac{x-3}{x^2+9}right)^2 , dx = | vedclass

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(C) We have $I = \int \left(\frac{x-3}{x^2+9}\right)^2 \, dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} \, dx$. Splitting the integral,we get $I = \int \fr

Vedclass · Accepted Answer

$\frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) + \frac{3}{x^2+9} + c$,where $c$ is the constant of integration.

Vedclass · Answer

(C) We have $I = \int \left(\frac{x-3}{x^2+9}ight)^2 \, dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} \, dx$. Splitting the integral,we get $I = \int \frac{x^2+9}{(x^2+9)^2} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$. $I = \int \frac{1}{x^2+9} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$. The first part is $\int \frac{1}{x^2+3^2} \, dx = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight)$. For the second part,let $u = x^2+9$,then $du = 2x \, dx$,so $3 \, du = 6x \, dx$. Thus,$\int \frac{6x}{(x^2+9)^2} \, dx = \int \frac{3}{u^2} \, du = 3 \left(-\frac{1}{u}ight) = -\frac{3}{x^2+9}$. Combining these,$I = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) - \left(-\frac{3}{x^2+9}ight) + c = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) + \frac{3}{x^2+9} + c$.

$\int \left(\frac{x-3}{x^2+9}\right)^2 \, dx =$

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