Question

\[ \int \frac{x}{1+x^4} d x= \]

• A. \(\frac{1}{2}\mathrm{tan}^{-1}x^{2}+c\), where \(c\) is the constant of integration
• B. \(2\mathrm{tan}^{-1}x+c\), where \(c\) is the constant of integration
• C. \(\frac{1}{2}\mathrm{tan}^{-1}x+c\), where c is the constant of integration
• D. \(\mathrm{tan}^{-1}x^{2}+\mathrm{c}\), where c is the constant of integration

Hint:

Substitute t = x squared so the integrand becomes the standard inverse tangent form.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Since the numerator has \(x\,dx\) and the denominator has \(x^4=(x^2)^2\), a substitution \(t=x^2\) works.

Step 2: Substitute
Put \(t=x^2\), so \(dt=2x\,dx\) and \(x\,dx=\dfrac{dt}{2}\).

Step 3: Calculation
\(\int\dfrac{x\,dx}{1+x^4}=\dfrac12\int\dfrac{dt}{1+t^2}=\dfrac12\tan^{-1}t+c=\dfrac12\tan^{-1}x^2+c\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The integral equals one-half times the inverse tangent of x squared, plus a constant. \[ \boxed{\text{(A) }\dfrac{1}{2}\tan^{-1}x^{2}+c} \]