Find the following integral: int frac{d x}{x^ | vedclass

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(A) We have $x^{2}-6 x+13 = x^{2}-6 x+9-9+13 = (x-3)^{2}+4 = (x-3)^{2}+2^{2}$.So,$\int \frac{d x}{x^{2}-6 x+13} = \int \frac{d x}{(x-3)^{2}+2^{2}}$.Le

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$\frac{1}{2} 	an^{-1} \left( \frac{x-3}{2} ight) + C$

Vedclass · Answer

(A) We have $x^{2}-6 x+13 = x^{2}-6 x+9-9+13 = (x-3)^{2}+4 = (x-3)^{2}+2^{2}$.So,$\int \frac{d x}{x^{2}-6 x+13} = \int \frac{d x}{(x-3)^{2}+2^{2}}$.Let $x-3 = t$. Then $dx = dt$.Using the standard integral formula $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) + C$,we get:$\int \frac{dt}{t^{2}+2^{2}} = \frac{1}{2} 	an^{-1} \left( \frac{t}{2} ight) + C$.Substituting $t = x-3$ back,we get:$= \frac{1}{2} 	an^{-1} \left( \frac{x-3}{2} ight) + C$.

Find the following integral: $\int \frac{d x}{x^{2}-6 x+13}$

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