int frac{1}{(x-2)(x^2+1)} dx= | vedclass

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(D) To solve the integral $\int \frac{1}{(x-2)(x^2+1)} dx$, we use partial fractions:$\frac{1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1}$$1 =

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$\frac{1}{5}\left[\log \frac{|x-2|}{\sqrt{1+x^2}}-2 	an ^{-1} xight]+c$

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(D) To solve the integral $\int \frac{1}{(x-2)(x^2+1)} dx$, we use partial fractions:$\frac{1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1}$$1 = A(x^2+1) + (Bx+C)(x-2)$Setting $x=2$, we get $1 = A(4+1) \Rightarrow 5A = 1 \Rightarrow A = \frac{1}{5}$.Comparing coefficients of $x^2$: $A+B = 0 \Rightarrow B = -A = -\frac{1}{5}$.Comparing constants: $A - 2C = 1 \Rightarrow \frac{1}{5} - 2C = 1 \Rightarrow -2C = \frac{4}{5} \Rightarrow C = -\frac{2}{5}$.Substituting these values into the integral:$\int \left( \frac{1/5}{x-2} + \frac{-1/5x - 2/5}{x^2+1} ight) dx = \frac{1}{5} \int \frac{1}{x-2} dx - \frac{1}{5} \int \frac{x}{x^2+1} dx - \frac{2}{5} \int \frac{1}{x^2+1} dx$$= \frac{1}{5} \log |x-2| - \frac{1}{5} \cdot \frac{1}{2} \log(x^2+1) - \frac{2}{5} 	an^{-1} x + C$$= \frac{1}{5} \left[ \log |x-2| - \frac{1}{2} \log(x^2+1) - 2 	an^{-1} x ight] + C$$= \frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{x^2+1}} - 2 	an^{-1} x ight] + C$.

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

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