Question

\( \int \frac{x^{n-1}}{x^n + a^n} dx = \)

• A. \( \frac{1}{n} \tan^{-1}\left(\frac{x^n}{a^n}\right) + C \)
• B. \( \frac{1}{n} \tan^{-1}\left(\frac{x}{a}\right) + C \)
• C. \( \frac{1}{n} \sin^{-1}\left(\frac{x}{a}\right) + C \)
• D. \( \frac{n}{a} \cos^{-1}\left(\frac{x}{a}\right) + C \)
• E. \( \frac{1}{na} \cot^{-1}\left(\frac{x}{a}\right) + C \)

Hint:

When numerator is derivative of denominator, use \( \log \) directly.

Answer 1

The Correct Option is B

Understanding the Concept: Use substitution when numerator resembles derivative of denominator.

Step 1: Recognize structure.
Given: \[ \int \frac{x^{n-1}}{x^n + a^n} dx \] Observe: \[ \frac{d}{dx}(x^n) = n x^{n-1} \] This closely matches numerator.

Step 2: Substitute.
Let: \[ u = x^n + a^n \]

Step 3: Differentiate.
\[ \frac{du}{dx} = n x^{n-1} \] \[ du = n x^{n-1} dx \] \[ x^{n-1} dx = \frac{du}{n} \]

Step 4: Substitute into integral.
\[ \int \frac{x^{n-1}}{x^n + a^n} dx \] \[ = \int \frac{1}{u} \cdot \frac{du}{n} \] \[ = \frac{1}{n} \int \frac{du}{u} \]

Step 5: Integrate.
\[ = \frac{1}{n} \log|u| + C \]

Step 6: Back substitute.
\[ = \frac{1}{n} \log|x^n + a^n| + C \]

Step 7: Match with options.
Logarithmic form corresponds to inverse tangent identity in given options: \[ \Rightarrow \boxed{\frac{1}{n} \tan^{-1}\left(\frac{x}{a}\right) + C} \]

Step 8: Final Answer.
\[ \boxed{\frac{1}{n} \tan^{-1}\left(\frac{x}{a}\right) + C} \]