int frac{d x}{xleft(x^2+1right)}= | vedclass

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(A) To evaluate the integral $I = \int \frac{dx}{x(x^2+1)}$,we use partial fractions. Let $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$. Mul

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$\log |x|-\frac{1}{2} \log \left(x^2+1\right)+c$,where $c$ is the constant of integration.

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(A) To evaluate the integral $I = \int \frac{dx}{x(x^2+1)}$,we use partial fractions. Let $\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$. Multiplying by $x(x^2+1)$,we get $1 = A(x^2+1) + (Bx+C)x$. $1 = (A+B)x^2 + Cx + A$. Comparing coefficients,we get $A=1$,$C=0$,and $A+B=0$,which implies $B=-1$. Thus,$\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}$. Integrating both sides,$I = \int \frac{1}{x} dx - \int \frac{x}{x^2+1} dx$. For the second integral,let $u = x^2+1$,then $du = 2x dx$,so $x dx = \frac{du}{2}$. $I = \log |x| - \frac{1}{2} \int \frac{du}{u} = \log |x| - \frac{1}{2} \log |x^2+1| + c$. Since $x^2+1 > 0$,we can write this as $\log |x| - \frac{1}{2} \log (x^2+1) + c$.

$\int \frac{d x}{x\left(x^2+1\right)}=$

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