Question

$$ \int \frac{dx}{x(x^{10}+1)} = $$

• A. $\frac{1}{10} \log \left( \frac{x^{10}}{x^{10}+1} \right) + c$
• B. $\frac{1}{10} \log \left( \frac{x^{10}}{x^{10}-1} \right) + c$
• C. $\log \left( \frac{x^{10}}{x^{10}+1} \right) + c$
• D. None

Hint:

General shortcut: $\int \frac{dx}{x(x^n+1)} = \frac{1}{n} \log \left( \frac{x^n}{x^n+1} \right) + c$. This works for any positive integer $n$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To integrate a rational function where the denominator contains a high power of $x$, we can manipulate the expression to create a substitution that involves the derivative of that high power.

Step 2: Key Formula or Approach:
Multiply both the numerator and the denominator by $x^9$ to make the substitution $u = x^{10}$ easier.

Step 3: Detailed Explanation:
1. Multiply by $x^9$: \[ \int \frac{x^9}{x^{10}(x^{10}+1)} \, dx \] 2. Let $u = x^{10} + 1$. Then $du = 10x^9 \, dx$, so $x^9 \, dx = \frac{du}{10}$. Also, $x^{10} = u - 1$. 3. Substitute into the integral: \[ \frac{1}{10} \int \frac{du}{(u-1)u} \] 4. Use partial fractions: $\frac{1}{u(u-1)} = \frac{1}{u-1} - \frac{1}{u}$. \[ \frac{1}{10} \int \left( \frac{1}{u-1} - \frac{1}{u} \right) \, du = \frac{1}{10} (\ln|u-1| - \ln|u|) + c \] 5. Substitute back $u = x^{10} + 1$: \[ \frac{1}{10} \ln \left| \frac{(x^{10}+1)-1}{x^{10}+1} \right| + c = \frac{1}{10} \log \left( \frac{x^{10}}{x^{10}+1} \right) + c \]

Step 4: Final Answer
The result of the integral is $\frac{1}{10} \log \left( \frac{x^{10}}{x^{10}+1} \right) + c$.