Q64 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics

$\int \dfrac{e^{\log\left(1+\frac{1}{x^2}\right)}}{x^2 + \dfrac{1}{x^2}}\ dx =$

  • Correct answer
    A
    $\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C$
  • B
    $\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C$
  • C
    $\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2+1}{x\sqrt{2}}\right) + C$
  • D
    $-\dfrac{1}{\sqrt{2}}\tan^{-1}\left(x - \dfrac{1}{x}\right) + C$

Solution

Given integral is $I = \int \dfrac{e^{\log\left(1+\dfrac{1}{x^2}\right)}}{x^2 + \dfrac{1}{x^2}}\ dx$

Using the property $e^{\log f(x)} = f(x)$, the integral simplifies to:

$I = \int \dfrac{1+\dfrac{1}{x^2}}{x^2 + \dfrac{1}{x^2}}\ dx$

The denominator can be rewritten by completing the square:

$x^2 + \dfrac{1}{x^2} = \left(x - \dfrac{1}{x}\right)^2 + 2$

So the integral becomes:

$I = \int \dfrac{1+\dfrac{1}{x^2}}{\left(x - \dfrac{1}{x}\right)^2 + 2}\ dx$

Let $x - \dfrac{1}{x} = t$

Differentiating both sides with respect to $x$, we get:

$\left(1 + \dfrac{1}{x^2}\right) dx = dt$

Substituting this into the integral:

$I = \int \dfrac{dt}{t^2 + (\sqrt{2})^2}$

Using the standard integral formula $\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C$:

$I = \dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{t}{\sqrt{2}}\right) + C$

Substituting back $t = x - \dfrac{1}{x} = \dfrac{x^2-1}{x}$:

$I = \dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C$

Answer: $\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^2-1}{\sqrt{2}\,x}\right) + C$

Hence, the correct answer is option: 'A'.

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