Question:
The integral $\int \frac{dx}{x(x^4 + 1)} =$
The integral $\int \frac{dx}{x(x^4 + 1)} =$
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For integrals of the form $\int \frac{dx}{x(x^n+1)}$, the result is always $\frac{1}{n}\ln\left|\frac{x^n}{x^n+1}\right| + C$.
Updated On: May 31, 2026
- $\frac{1}{4} \ln \left| \frac{x^4}{x^4 + 1} \right| + C$
- $\frac{1}{4} \ln \left| \frac{x^4 + 1}{x^4} \right| + C$
- $\ln \left| \frac{x^4}{x^4 + 1} \right| + C$
- $\ln \left| \frac{x^4 + 1}{x^4} \right| + C$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
We use substitution by manipulating the integrand. Multiplying both the numerator and the denominator by $x^3$ allows us to substitute $u = x^4$.
Step 2: Meaning
We can rewrite the integrand as: \[ I = \int \frac{x^3}{x^4(x^4 + 1)} \, dx \]
Step 3: Analysis
Let: \[ u = x^4 \implies du = 4x^3 \, dx \implies x^3 \, dx = \frac{du}{4} \] Substitute this into the integral: \[ I = \frac{1}{4} \int \frac{du}{u(u + 1)} \] Using partial fractions: \[ I = \frac{1}{4} \int \left(\frac{1}{u} - \frac{1}{u+1}\right) \, du \] \[ I = \frac{1}{4} \left( \ln|u| - \ln|u+1| \right) + C = \frac{1}{4} \ln \left| \frac{u}{u+1} \right| + C \] Substituting $u = x^4$ back into the expression: \[ I = \frac{1}{4} \ln \left| \frac{x^4}{x^4 + 1} \right| + C \]
Step 4: Conclusion
The value of the indefinite integral is $\frac{1}{4} \ln \left| \frac{x^4}{x^4 + 1} \right| + C$. Final Answer: (A)
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