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

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

Vedclass · Accepted Answer

$\log |x|-\frac{1}{2} \log (x^{2}+1)+C$

Vedclass · Answer

(A) To evaluate the integral $\int \frac{dx}{x(x^{2}+1)}$,we use partial fractions.Let $\frac{1}{x(x^{2}+1)} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1}$.Multiplying both sides by $x(x^{2}+1)$,we get $1 = A(x^{2}+1) + (Bx+C)x$.Comparing coefficients of $x^{2}$,$x$,and the constant term,we get $A+B=0$,$C=0$,and $A=1$.Thus,$A=1$,$B=-1$,and $C=0$.So,$\frac{1}{x(x^{2}+1)} = \frac{1}{x} - \frac{x}{x^{2}+1}$.Integrating both sides,we get $\int \frac{dx}{x} - \int \frac{x}{x^{2}+1} dx$.$= \log |x| - \frac{1}{2} \log |x^{2}+1| + C$.Since $x^{2}+1 > 0$ for all real $x$,we can write this as $\log |x| - \frac{1}{2} \log (x^{2}+1) + C$.

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

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