Question

The integral \[ \int \frac{x^2}{1+x^6}\,dx \] is equal to:

• A. $x^{3} + C$
• B. $\frac{1}{3} \text{Tan}^{-1}(x^{3}) + C$
• C. $\log(1+x^{3})$
• D. $\frac{1}{1+x^{3}} + C$

Hint:

Whenever the power in the denominator is twice the power in the numerator plus one (excluding the constant), consider an inverse trigonometric substitution.

Answer 1

The Correct Option is B

Step 1: Concept
This integral can be solved using the substitution method to transform it into a standard form. The standard integral we use is $\int \frac{1}{1+u^{2}} du = \text{Tan}^{-1}(u) + C$.

Step 2: Meaning
We can rewrite the denominator $x^{6}$ as $(x^{3})^{2}$. This suggests that substituting $u = x^{3}$ will simplify the numerator $x^{2} dx$.

Step 3: Analysis
Let $u = x^{3}$. Then, differentiating both sides gives $du = 3x^{2} dx$, which implies $x^{2} dx = \frac{1}{3} du$. Substituting these into the original integral: $\int \frac{x^{2}}{1+(x^{3})^{2}} dx = \int \frac{1}{1+u^{2}} \left( \frac{1}{3} du \right)$ $= \frac{1}{3} \int \frac{1}{1+u^{2}} du$ $= \frac{1}{3} \text{Tan}^{-1}(u) + C$.

Step 4: Conclusion
Replacing $u$ with the original variable $x^{3}$, we get $\frac{1}{3} \text{Tan}^{-1}(x^{3}) + C$. Final Answer: (B)