Question

\( \int \frac{x}{1+x^4} dx = \)}

• A. \( \frac{1}{2} \tan^{-1} x^2 + c \)
• B. \( 2 \tan^{-1} x + c \)
• C. \( \frac{1}{2} \tan^{-1} x + c \)
• D. \( \tan^{-1} x^2 + c \)

Hint:

Whenever the numerator power is $(n-1)$ and denominator power is $(2n)$, substitute $u = x^n$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to evaluate the indefinite integral of a rational function.
The integrand is \( \frac{x}{1+x^4} \).
We observe that the derivative of \( x^2 \) is \( 2x \), which is present in the numerator (after adjusting the constant).
Step 2: Key Formula or Approach:
We use the method of substitution and the standard integral formula:
\[ \int \frac{1}{1+u^2} du = \tan^{-1} u + c \] Step 3: Detailed Explanation:
Let \( u = x^2 \).
Differentiating both sides with respect to \( x \), we get:
\[ du = 2x dx \implies x dx = \frac{1}{2} du \] Substituting these into the original integral:
\[ \int \frac{x}{1+x^4} dx = \int \frac{1}{1+(x^2)^2} (x dx) \] \[ = \int \frac{1}{1+u^2} \left( \frac{1}{2} du \right) \] \[ = \frac{1}{2} \int \frac{1}{1+u^2} du \] \[ = \frac{1}{2} \tan^{-1} u + c \] Substituting \( u = x^2 \) back into the result:
\[ = \frac{1}{2} \tan^{-1} x^2 + c \] Step 4: Final Answer:
The evaluated integral is \( \frac{1}{2} \tan^{-1} x^2 + c \).