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  • Integrate the functions in Exercises 1 to 24. 4

Integrate the functions in Exercises 1 to 24.

    Q4.    \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}

Answers (1)

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For the simplifying the expression, we will multiply and dividing it by  x-3.

We then have,

                                            \frac{x^{-3}}{x^2 x^{-3}(x^4 + 1)^\frac{3}{4}}\ =\ \frac{1}{x^5}\left [ \frac{x^4\ +\ 1}{x^4} \right ]^{\frac{-3}{4}}

Now, let   

                                        \frac{1}{x^4}\ =\ t\ \Rightarrow \ \frac{1}{x^5}dx\ =\ \frac{-dt}{4}

Thus,                               

                                         \int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx

or                                                                         =\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt

                                                                             =\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C

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

Posted by

Devendra Khairwa

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