If int frac{d x}{(x+2)left(x^2+1right)}=a log | vedclass

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(A) Given,$\int \frac{d x}{(x+2)\left(x^2+1ight)}=a \log \left|1+x^2ight|+b 	an ^{-1} x+\frac{1}{5} \log |x+2|+c$. Let $\frac{1}{(x+2)\left(x^2+1

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$a=\frac{-1}{10}, b=\frac{2}{5}$

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(A) Given,$\int \frac{d x}{(x+2)\left(x^2+1ight)}=a \log \left|1+x^2ight|+b 	an ^{-1} x+\frac{1}{5} \log |x+2|+c$. Let $\frac{1}{(x+2)\left(x^2+1ight)}=\frac{A}{(x+2)}+\frac{B x+C}{\left(x^2+1ight)}$. Then $1=A(x^2+1)+(x+2)(Bx+C) = (A+B)x^2 + (2B+C)x + (A+2C)$. Comparing coefficients,we get $A+B=0$,$2B+C=0$,and $A+2C=1$. Solving these,we find $A=\frac{1}{5}$,$B=-\frac{1}{5}$,and $C=\frac{2}{5}$. Thus,$\int \frac{d x}{(x+2)\left(x^2+1ight)} = \int \left( \frac{1}{5(x+2)} - \frac{x}{5(x^2+1)} + \frac{2}{5(x^2+1)} ight) dx$. $= \frac{1}{5} \log |x+2| - \frac{1}{10} \log |x^2+1| + \frac{2}{5} 	an^{-1} x + c$. Comparing this with the given expression,we get $a=-\frac{1}{10}$ and $b=\frac{2}{5}$.

If $\int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$,then

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