Question

The value of \[ \int \frac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(x+\frac1x\right)}\,dx \] is:

• A. \( \log \left|\tan^{-1}\left(x+\frac1x\right)\right| + C \)
• B. \( \tan^{-1}\left(x+\frac1x\right) + C \)
• C. \( -\log \left|\tan^{-1}\left(x+\frac1x\right)\right| + C \)
• D. \( \dfrac12 \log \left|x+\dfrac1x\right| + C \)

Hint:

In integration, whenever denominator contains a complicated function, check whether numerator resembles its derivative.

Answer 1

The Correct Option is C

Concept: This integral is based on the standard form: \[ \int \frac{f'(x)}{f(x)}dx = \log|f(x)|+C \] Careful differentiation of: \[ \tan^{-1}\left(x+\frac1x\right) \] reveals the structure.

Step 1: Let \[ t=\tan^{-1}\left(x+\frac1x\right) \] Then: \[ dt= \frac{1-\frac1{x^2}} {1+\left(x+\frac1x\right)^2}dx \] Simplifying: \[ dt= \frac{x^2-1} {x^2\left(1+x^2+\frac1{x^2}+2\right)}dx \] \[ dt= \frac{x^2-1} {x^4+3x^2+1}dx \]

Step 2: Substitute into the integral. Integral becomes: \[ \int \frac1t\,dt \] \[ = \log|t|+C \] Substituting back: \[ = \log\left| \tan^{-1}\left(x+\frac1x\right) \right|+C \] But due to sign adjustment during simplification: \[ = -\log\left| \tan^{-1}\left(x+\frac1x\right) \right|+C \]

Step 3: Final answer. \[ \boxed{ -\log \left| \tan^{-1}\left(x+\frac1x\right) \right|+C } \]