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The integral  \int \frac{dx}{x^{2}(x^{4}+1)^{3/4}}\; equals\, :

  • Option 1)

    \left [ \frac{x^{4}+1}{x^{4}} \right ]^{\frac{1}{4}}+c\;

  • Option 2)

    (x^{4}+1)^{\frac{1}{4}}+c\;

  • Option 3)

    -(x^{4}+1)^{\frac{1}{4}}+c\;

  • Option 4)

    -\left [ \frac{x^{4}+1}{x^{4}} \right ]^{\frac{1}{4}}+c

 

Answers (1)

As learnt in concept

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 )

 

 

\int \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{2}}}

=\int \frac{dx}{x^{2}\times (x^{4})^{\frac{3}{4}}(1+\frac{1}{x^{4}})^{\frac{3}{4}}}

=>\int \frac{dx}{x^{5}}\times \frac{1}{(1+\frac{1}{x^{4}})^{\frac{3}{4}}}

If 1+\frac{1}{x^{4}}=t

\frac{-4}{x^{5}}dx=dt

=-\frac{1}{4}\int dt\times \frac{1}{t^{\frac{3}{4}}} 

=\frac{-1}{4}\left [ \frac{t^{\frac{1}{4}}}{\frac{1}{4}} \right ]+C

=-\left [ \frac{x^{4}+1}{x^{4}} \right ]^{\frac{1}{4}}+C

 


Option 1)

\left [ \frac{x^{4}+1}{x^{4}} \right ]^{\frac{1}{4}}+c\;

Option 2)

(x^{4}+1)^{\frac{1}{4}}+c\;

Option 3)

-(x^{4}+1)^{\frac{1}{4}}+c\;

Option 4)

-\left [ \frac{x^{4}+1}{x^{4}} \right ]^{\frac{1}{4}}+c

Posted by

Vakul

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