INTEGRALS — Maths STD 12 Science — Question

Multipling and dividing by  $\text{x}^{-3}$, we obtain

$\frac{\text{x}^{-3}}{\text{x}^{2}.\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{2}}}{\text{x}^{2}.\text{x}^{-3}}$

$=\frac{(\text{x}^{4}+1)^{\frac{-3}{4}}}{\text{x}^{5}.(\text{x)}^{4^{-\frac{3}{4}}}}$

$=\frac{1}{\text{x}^{5}}\bigg(\frac{\text{x}^{4}+1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}$

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

$\therefore\int\frac{1}{\text{x}^{2}(\text{x}^{4}+1)^{\frac{3}{4}}}\text{dx}=\int\frac{1}{\text{x}^{5}}\bigg(1+\frac{1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}\text{dx}$

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

$=-\frac{1}{4}\Bigg[\frac{(1+\text{t)}^{-\frac{3}{4}}}{\frac{1}{4}}\Bigg]+\text{C}$

$=-\frac{1}{4}\frac{\bigg(1+\frac{1}{\text{x}^{4}}\Bigg)^{\frac{1}{4}}}{\frac{1}{4}}+\text{C}$

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