Question

The value of the integral \(\int \frac{dx}{x^{4}+x^{3}}\) is

• A. \(-\frac{1}{2x^{2}}+\frac{1}{x}+\log |\frac{x}{x+1}|+c\)
• B. \(\frac{1}{2x^{2}}+\frac{1}{x}+\log |\frac{x}{x+1}|+c\)
• C. \(-\frac{1}{2x^{2}}-\frac{1}{x}+\log |\frac{x}{x+1}|+c\)
• D. \(-\frac{1}{2x^{2}}-\frac{1}{x}-\log |\frac{x}{x+1}|+c\)

Hint:

Factor the denominator as x cubed times (x + 1) and split into partial fractions.

Answer 1

The Correct Option is A

Step 1: Differentiate Option (A)
Differentiate $F=-\dfrac{1}{2x^{2}}+\dfrac{1}{x}+\log x-\log(x+1)$.

Step 2: Compute
$F'=\dfrac{1}{x^{3}}-\dfrac{1}{x^{2}}+\dfrac{1}{x}-\dfrac{1}{x+1}$.

Step 3: Combine
Over the common denominator $x^{3}(x+1)$ the numerator is $1$, so $F'=\dfrac{1}{x^{3}(x+1)}=\dfrac{1}{x^{4}+x^{3}}$. Option (A) is correct.

Final Answer:
Option (A) differentiates back to the integrand. \[ \boxed{\text{(A) }-\dfrac{1}{2x^{2}}+\dfrac{1}{x}+\log \left|\dfrac{x}{x+1}\right|+c} \]