Question

\(\int \frac{\mathrm{d} x}{x\left(x^3+1\right)}=\)

• A. \(\mathrm{log}(\frac{x^{3}}{x^{3}+1})+\mathrm{c}\), where c is the constant of integration
• B. \(\frac{1}{3}\mathrm{log}(\sqrt[\frac{x^{3}}{x^{3}+1}]{3})+\mathrm{c}\), where c is the constant of integration
• C. \(\mathrm{log}(\sqrt[\frac{x^{3}}{x^{3}+1}]{3})+\mathrm{c}\), where c is the constant of integration
• D. \(\frac{1}{3}\mathrm{log}(\frac{x^{3}+1}{x^{3}})+\mathrm{c}\), where c is the constant of integration

Hint:

Multiply the numerator and denominator by x squared and substitute t = x cubed.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Write \(\dfrac{1}{x(x^3+1)}=\dfrac{x^2}{x^3(x^3+1)}\).

Step 2: Substitute
Put \(t=x^3\), so \(x^2dx=\dfrac{dt}{3}\). The integral is \(\dfrac13\int\dfrac{dt}{t(t+1)}\).

Step 3: Partial fractions
\(\dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1}\), so the integral is \(\dfrac13\left[\log t-\log(t+1)\right]=\dfrac13\log\dfrac{t}{t+1}\).

Step 4: Compare the options
The result \(\dfrac13\log\dfrac{x^3}{x^3+1}+c\) equals \(\log\left(\dfrac{x^3}{x^3+1}\right)^{1/3}+c\), which is option (C) (the cube root form).

Final Answer:
The integral equals one-third of the log of x cubed over (x cubed + 1), plus c. \[ \boxed{\text{(C) }\log\left(\dfrac{x^{3}}{x^{3}+1}\right)^{\frac13}+c} \]