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The value of the integral \int \left ( x+\frac{1}{x} \right )^{n+5}\left ( \frac{x^{2}-1}{x^{2}} \right )dx is equal to

  • Option 1)

    \frac{\left ( x+\frac{1}{x} \right )^{n+6}}{n+6}+c

  • Option 2)

    \left ( \frac{ x^{2}+1}{x^{2}}\right )^{n+6}\left ( n+6 \right )+c

  • Option 3)

    \left ( \frac{ x}{x^{2}+1}\right )^{n+6}\left ( n+6 \right )+c

  • Option 4)

    none

 

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

Integration in the form of 

(i) \int f(x+\frac{1}{x})(1-\frac{1}{x^{2}})dx

(ii) \int f(x-\frac{1}{x})(1+\frac{1}{x^{2}})dx

(iii) \int f(x^{2}+\frac{1}{x^{2}})(x-\frac{1}{x^{3}})dx

(iv) \int f(x^{2}-\frac{1}{x^{2}})(x+\frac{1}{x^{3}})dx

(v) \int \frac{(1\pm \frac{1}{x^{2}})dx}{x^{2}+\frac{1}{x^{2}}}

(vi) \int \frac{f(x)dx}{ax^{4}+2bx^{3}+cx^{2}+2bx+a}  

- wherein

(i) \rightarrow put   (x+\frac{1}{x})=t

(ii) \rightarrow put    (x-\frac{1}{x})=t

(iii) \rightarrow put  (x^{2}+\frac{1}{x^{2}})=t

(iv) \rightarrow put (x^{2}-\frac{1}{x^{2}})=t

(v) \rightarrow for 1+\frac{1}{x^{2}}   put    x-\frac{1}{x}=t

     \rightarrow for 1-\frac{1}{x^{2}}   put    x+\frac{1}{x}=t 

(vi) \rightarrow put (x+\frac{1}{x})=t  if  b\neq 0

            put(x^{2}+\frac{1}{x^{2}})=t if b= 0

 

 

 I = \int {{p^{n + 5}}} dp\: \: \: \: If\: \: x + \frac{1}{x} = p\: \: then,\left( {1 - \frac{1}{{{x^2}}}} \right)\,dx = dp

\therefore I = \int {{{\left( {x + \frac{1}{x}} \right)}^{n + 3}}} \;\left( {\frac{{{x^2} - 1}}{{{x^2}}}} \right)\;dx = \int {{p^{n + 5}}} dp = \frac{{{p^{n + 6}}}}{{n + 6}} + c = \frac{{{{\left( {x + \frac{1}{x}} \right)}^{n + 6}}}}{{n + 6}} + c


Option 1)

\frac{\left ( x+\frac{1}{x} \right )^{n+6}}{n+6}+c

Option 2)

\left ( \frac{ x^{2}+1}{x^{2}}\right )^{n+6}\left ( n+6 \right )+c

Option 3)

\left ( \frac{ x}{x^{2}+1}\right )^{n+6}\left ( n+6 \right )+c

Option 4)

none

Posted by

gaurav

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