Question

The value of \(\int\frac{x^2dx}{\sqrt{x^6+a^6}}\) is equal to

• A. \(\log|x^3+\sqrt{x^6+a^6}|+c\)
• B. \(\log|x^3-\sqrt{x^6+a^6}|+c\)
• C. \(\frac{1}{3}\log|x^3+\sqrt{x^6+a^6}|+c\)
• D. \(\frac{1}{3}\log|x^3-\sqrt{x^6+a^6}|+c\)

Answer 1

The Correct Option is D

The given integral is: \[ I = \int \frac{x^2}{\sqrt{x^6 + a^6}} \, dx \] To solve this, use the substitution \( u = x^3 \), so that \( du = 3x^2 \, dx \). The integral becomes: \[ I = \frac{1}{3} \int \frac{du}{\sqrt{u^2 + a^6}} \] This is a standard integral, and its solution is: \[ \frac{1}{3} \log | u + \sqrt{u^2 + a^6} | + c \] Substituting \( u = x^3 \) back into the result, we get: \[ I = \frac{1}{3} \log | x^3 + \sqrt{x^6 + a^6} | + c \] Therefore, the correct answer is (D).

Answer 2

Given Integral:
$$ \int \frac{x^2 \, dx}{\sqrt{x^6 + a^6}} $$ 
Step 1: Substitution
Let: $$ u = x^3 \Rightarrow du = 3x^2 \, dx \Rightarrow \frac{du}{3} = x^2 \, dx $$ 
Step 2: Rewrite the integral in terms of u
Then the integral becomes: $$ \int \frac{x^2 \, dx}{\sqrt{x^6 + a^6}} = \int \frac{1}{\sqrt{u^2 + a^6}} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{du}{\sqrt{u^2 + a^6}} $$ 
Step 3: Apply standard formula
We use the standard integral: $$ \int \frac{du}{\sqrt{u^2 + k^2}} = \ln|u + \sqrt{u^2 + k^2}| + C $$ So, $$ \frac{1}{3} \ln|u + \sqrt{u^2 + a^6}| + C $$ 
Step 4: Back-substitute
Since $u = x^3$: $$ \frac{1}{3} \ln|x^3 + \sqrt{x^6 + a^6}| + C $$ 
Final Answer:
$\frac{1}{3} \log|x^3 + \sqrt{x^6 + a^6}| + C$