Question

Evaluate the integral \[ \int \frac{x^2+1}{x^4+x^2+1}\,dx \]

• A. \( \frac{1}{\sqrt{3}}\tan^{-1}\!\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c \)
• B. \( \frac{1}{3}\tan^{-1}\!\left(\frac{x-\frac{1}{x}}{3}\right)+c \)
• C. \( \frac{1}{\sqrt{3}}\tan^{-1}\!\left(\frac{x+\frac{1}{x}}{\sqrt{3}}\right)+c \)
• D. \( \frac{1}{3}\tan^{-1}\!\left(\frac{x+\frac{1}{x}}{3}\right)+c \)

Hint:

When expressions involve \(x\) and \(\frac{1}{x}\), try substitutions like \(x-\frac{1}{x}\) or \(x+\frac{1}{x}\).

Answer 1

The Correct Option is A

Step 1: Divide numerator and denominator by \( x^2 \).
\[ \int \frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\,dx \]

Step 2: Use substitution.
Let \[ t = x - \frac{1}{x} \Rightarrow dt = \left(1+\frac{1}{x^2}\right)dx \]

Step 3: Simplify the denominator.
\[ x^2 + 1 + \frac{1}{x^2} = t^2 + 3 \]

Step 4: Integrate.
\[ \int \frac{dt}{t^2+3} = \frac{1}{\sqrt{3}}\tan^{-1}\!\left(\frac{t}{\sqrt{3}}\right)+c \]

Step 5: Back substitute.
\[ \boxed{\frac{1}{\sqrt{3}}\tan^{-1}\!\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+c} \]