Integrate the function in Exercise: 1 x^ 2 (x^ 2 +1)^ 3 4 - Vidyadip

Question

Integrate the function in Exercise:
$\frac{1}{\text{x}^{2}(\text{x}^{2}+1)^{\frac{3}{4}}}$

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Answer

$\frac{1}{\text{x}^{2}(\text{x}^{2}+1)^{\frac{3}{4}}}$Multipling and dividing by $\text{x}^{-3}$, we obtain
$\frac{\text{x}^{-3}}{\text{x}^{2}.\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{2}}}{\text{x}^{2}.\text{x}^{-3}}$
$=\frac{(\text{x}^{4}+1)^{\frac{-3}{4}}}{\text{x}^{5}.(\text{x)}^{4^{-\frac{3}{4}}}}$
$=\frac{1}{\text{x}^{5}}\bigg(\frac{\text{x}^{4}+1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}$
$\text{Let}\frac{1}{\text{x}^{4}}=\text{t} \Rightarrow-\frac{4}{\text{x}^{5}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{5}}\text{dx}=-\frac{\text{dt}}{4}$
$\therefore\int\frac{1}{\text{x}^{2}(\text{x}^{4}+1)^{\frac{3}{4}}}\text{dx}=\int\frac{1}{\text{x}^{5}}\bigg(1+\frac{1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}\text{dx}$
$=-\frac{1}{4}\int(1+\text{t)}^{-\frac{3}{4}}\text{dt}$
$=-\frac{1}{4}\Bigg[\frac{(1+\text{t)}^{-\frac{3}{4}}}{\frac{1}{4}}\Bigg]+\text{C}$
$=-\frac{1}{4}\frac{\bigg(1+\frac{1}{\text{x}^{4}}\Bigg)^{\frac{1}{4}}}{\frac{1}{4}}+\text{C}$
$=-\Bigg(1+\frac{1}{\text{x}^{4}}\Bigg)^{\frac{1}{4}}+\text{C}$

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