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  • Please,please help me - The integral is equal to : (where C is a constant of integration) - Integral Calculus - JEE Main

 

The integral \int \frac{3x^{13}+2x^{11}}{\left ( 2x^{4}+3x^{2}+1 \right )^{4}}dx  is equal to : 

(where C is a constant of integration) 

  • Option 1)

    \frac{x^{4}}{\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

     

     

     

  • Option 2)

    \frac{x^{4}}{6\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

  • Option 3)

     

    \frac{x^{12}}{\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

  • Option 4)

     

    \frac{x^{12}}{6\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

Answers (1)

best_answer

 

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

 

Type of integration by substitution -

\int \frac{f'(x)dx}{\sqrt{f(x)}}=2\sqrt{f(x)}+c

- wherein

Let f(x)=t 

\therefore f{}'(x)dx=dt

 

I=\int \frac{(3x^{13}+2x^{11})dx}{(4x^{4}+3x^{2}+1)^{4}}=\int \frac{(3x^{13}+2x^{11})dx}{x^{16}(4+3x^{-2}+1^{-4})^{4}}\\\\=\int \frac{3x^{-3}+2x^{-5}}{(4+3x^{-2}+x^{-4})^{4}}dx\\\\\\put \: 4+3x^{-2}+x^{-4}=t\\\\\\dt=-2(3x^{-3}+2x^{-5})dx\\\\\therefore I=-\frac{1}{2}\int \frac{dt}{t^{4}}=-\frac{1}{2}(\frac{t^{-3}}{-3})+c\\\\=\frac{1}{6(4+3x^{-2}+x^{-4})^{3}}+c\\\\=\frac{x^{12}}{6(4x^{4}+3x^{2}+1)^{3}}+c

 


Option 1)

\frac{x^{4}}{\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

 

 

 

Option 2)

\frac{x^{4}}{6\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

Option 3)

 

\frac{x^{12}}{\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C

Option 4)

 

\frac{x^{12}}{6\left ( 2x^{4}+3x^{2}+1 \right )^{3}}+C
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