Question

$\int \frac{dx}{x(x^3 + 1)} = \dots$

• A. $\log \left(\frac{x^3}{x^3 + 1}\right) + c$
• B. $\frac{1}{3} \log \sqrt{\frac{x^3 + 1}{x^3}} + c$
• C. $\log \sqrt{\frac{x^3 + 1}{x^3}} + c$
• D. $\frac{1}{3} \log \left(\frac{x^3}{x^3 + 1}\right) + c$

Hint:

Memorize this highly recurring integral shortcut: $\int \frac{dx}{x(x^n + 1)} = \frac{1}{n} \log \left| \frac{x^n}{x^n + 1} \right| + c$. You can apply it instantly to save time.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are tasked with evaluating a rational indefinite integral. The denominator contains a higher power of $x$ mixed with a linear $x$.

Step 2: Key Formula or Approach:
For integrals of the form $\int \frac{dx}{x(x^n + 1)}$, the standard algebraic trick is to multiply both the numerator and the denominator by $x^{n-1}$.
This perfectly sets up the integral for a straightforward substitution of $t = x^n$.

Step 3: Detailed Explanation:
Here, $n=3$. We multiply the numerator and denominator by $x^2$:
$$\int \frac{x^2 \, dx}{x^3(x^3 + 1)}$$
Now, use the substitution $t = x^3$.
Differentiate with respect to $x$: $dt = 3x^2 \, dx \implies \frac{dt}{3} = x^2 \, dx$.
Substitute these into the integral:
$$= \frac{1}{3} \int \frac{dt}{t(t + 1)}$$
Use partial fraction decomposition on $\frac{1}{t(t+1)}$:
$$\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1} \implies 1 = A(t+1) + B(t)$$
Setting $t=0$ gives $A=1$. Setting $t=-1$ gives $B=-1$.
So, the integral breaks down into:
$$= \frac{1}{3} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$$
Integrate the terms:
$$= \frac{1}{3} \left( \log|t| - \log|t+1| \right) + c$$
Apply logarithm quotient properties:
$$= \frac{1}{3} \log \left| \frac{t}{t+1} \right| + c$$
Back-substitute $t = x^3$:
$$= \frac{1}{3} \log \left| \frac{x^3}{x^3 + 1} \right| + c$$

Step 4: Final Answer:
The answer matches option (d).