Question

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

• A. \(\mathrm{log}(x)-\frac{1}{2}\mathrm{log}(x^{2}+1)+\mathrm{c}\), where c is the constant of integration.
• B. \(\frac{1}{2}\mathrm{log}(x)-\mathrm{log}(x^{2}+1)+\mathrm{c}\), where c is the constant of integration.
• C. \(\mathrm{log}(x)+\frac{1}{2}\mathrm{log}(x^{2}+1)+\mathrm{c}\), where c is the constant of integration.
• D. \(-\mathrm{log}(x)-\frac{1}{2}\mathrm{log}(x^{2}+1)+\mathrm{c}\), where c is the constant of integration.

Hint:

Split the fraction into partial fractions and integrate each part.

Answer 1

The Correct Option is A

Step 1: Rewrite the Integrand
$\dfrac{1}{x(x^{2}+1)}=\dfrac{(x^{2}+1)-x^{2}}{x(x^{2}+1)}=\dfrac{1}{x}-\dfrac{x}{x^{2}+1}$.

Step 2: Integrate Term by Term
$\int\dfrac{dx}{x}=\log x$ and $\int\dfrac{x\,dx}{x^{2}+1}=\dfrac{1}{2}\log(x^{2}+1)$.

Step 3: Result
The answer is $\log x-\dfrac{1}{2}\log(x^{2}+1)+c$, option (A).

Final Answer:
The result is $\log x-\dfrac{1}{2}\log(x^{2}+1)$ plus a constant. \[ \boxed{\text{(A) }\log x-\dfrac{1}{2}\log\left(x^{2}+1\right)+c} \]