Question

$\int\frac{x\ dx}{x(x^2 +1)}$ equals
(A)$\log \arrowvert x \arrowvert - \frac12 \ log(x^2 +1) + C$
(B)$\log \arrowvert x \arrowvert + \frac12 \ log(x^2 +1) + C$
(C)$-\log \arrowvert x \arrowvert + \frac12 \ log(x^2 +1) + C$
(D)$\frac 12 \log \arrowvert x \arrowvert + \frac12 \ log(x^2 +1) + C$

Answer 1

Answer:

    Option (A) is correct.

Solution:

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

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

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

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

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

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

where C is constant of integration

Therefore, the required integral is

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

Rule:

    $\int \frac{dx}{x}=log|x|+C$

where C is constant of integration