Question

\(\int \frac{1}{x(1 + x^{4})} \, dx\) is equal to

• A. \(\frac{1}{4} \log_{e}\left|\frac{x^4}{1 + x^{4}}\right| + C\)
• B. \(\frac{1}{5} \log_{e}\left|\frac{x}{1 + x^{4}}\right| + C\)
• C. \(\frac{1}{4} \log_{e}\left|\frac{x}{1 + x^{4}}\right| + C\)
• D. \(\frac{1}{2} \log_{e}\left|\frac{x^2}{1 + x^{4}}\right| + C\)
• E. \(\log_{e}\left|\frac{x}{1 + x^{4}}\right| + C\)

Hint:

For integrals of the form \(\int \frac{dx}{x(1+x^n)}\), use substitution \(t = x^n\).

Answer 1

The Correct Option is A

Step 1: Concept:
• Use substitution: \[ t = x^4 \]

Step 2: Detailed Explanation:
• Given integral: \[ \int \frac{1}{x(1+x^4)} \, dx \]
• Rewrite: \[ = \int \frac{x^3}{x^4(1+x^4)} \, dx \]
• Let: \[ t = x^4 \Rightarrow dt = 4x^3 dx \Rightarrow x^3 dx = \frac{dt}{4} \]
• Substitute: \[ \int \frac{x^3}{x^4(1+x^4)} \, dx = \int \frac{1}{t(1+t)} \cdot \frac{dt}{4} \]
• Split using partial fractions: \[ = \frac{1}{4} \int \left(\frac{1}{t} - \frac{1}{1+t}\right) dt \]
• Integrate: \[ = \frac{1}{4} [\ln|t| - \ln|1+t|] + C \]
• Simplify: \[ = \frac{1}{4} \ln\left|\frac{t}{1+t}\right| + C = \frac{1}{4} \ln\left|\frac{x^4}{1+x^4}\right| + C \]

Step 3: Final Answer:
• \[ \frac{1}{4} \log_{e}\left|\frac{x^4}{1 + x^{4}}\right| + C \]