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\int \frac{\left ( x^{2} +1\right )}{\left ( x^{4} -x^{2}+1\right )cot^{-1}\left ( x-\frac{1}{x} \right ) }dx=

  • Option 1)

    -ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

  • Option 2)

    ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

  • Option 3)

    ln\left |x^{2} cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

  • Option 4)

    x^{2}ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

 

Answers (1)

best_answer

As we learnt

Special type of indefinite integration -

Integration of the form : g(sin^{-1}f(x))

Put f(x)=t 

so f'(x)dx=dt

Now Put \Theta =sin^{-1}t

such that t=sin\Theta

-

 

 Let us solve this question by analyzing the options given in the answer.

\frac{d}{{dx}}\left( { - \ln \left| {{{\cot }^{ - 1}}\left( {x - \frac{1}{x}} \right)} \right| + C} \right) = \frac{{ - 1}}{{{{\cot }^{ - 1}}\left( {x - \frac{1}{x}} \right)}}.\left( {\frac{{ - 1}}{{1 + {{\left( {x - \frac{1}{x}} \right)}^2}}}} \right).\left( {1 + \frac{1}{{{x^2}}}} \right)

 


Option 1)

-ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

Option 2)

ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

Option 3)

ln\left |x^{2} cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

Option 4)

x^{2}ln\left | cot^{-1}\left ( x-\frac{1}{x} \right ) \right |+C

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