Question

If \(I=\displaystyle\int_0^1 \frac{x^2}{1+x^6}\,dx\), then \(I=\)

• A. \(\frac{\pi}{8}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{12}\)
• D. \(\frac{\pi}{2}\)

Hint:

When an integral contains \(x^2dx\) and \(1+x^6\), use the substitution \(u=x^3\).

Answer 1

The Correct Option is C

Concept:
The integral is: \[ I=\int_0^1 \frac{x^2}{1+x^6}\,dx \] Since: \[ x^6=(x^3)^2 \] and the numerator contains: \[ x^2dx \] we use the substitution: \[ u=x^3 \]

Step 1: Use substitution.
Let: \[ u=x^3 \] Differentiate: \[ du=3x^2dx \] So: \[ x^2dx=\frac{du}{3} \]

Step 2: Change the limits.
When: \[ x=0 \] \[ u=0^3=0 \] When: \[ x=1 \] \[ u=1^3=1 \]

Step 3: Rewrite the integral.
\[ I=\int_0^1 \frac{x^2}{1+x^6}\,dx \] Since: \[ x^6=u^2 \] and: \[ x^2dx=\frac{du}{3} \] Therefore: \[ I=\frac{1}{3}\int_0^1 \frac{du}{1+u^2} \]

Step 4: Integrate.
We know: \[ \int \frac{du}{1+u^2}=\tan^{-1}u \] So: \[ I=\frac{1}{3}\left[\tan^{-1}u\right]_0^1 \] \[ I=\frac{1}{3}\left(\tan^{-1}1-\tan^{-1}0\right) \] \[ I=\frac{1}{3}\left(\frac{\pi}{4}-0\right) \] \[ I=\frac{\pi}{12} \] Hence, the correct answer is: \[ \boxed{(C)\ \frac{\pi}{12}} \]