Question
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
✓
Answer
$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Multiplying and dividing by $x^{-3},$ we obtain
$\frac{\text{x}^{-3}}{\text{x}^2.\text{x}^{-3}(\text{x}^4+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^2.\text{x}^{-3}}$
$=\frac{(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^5.(\text{x}^{4})^{-\frac{3}{4}}}$
$=\frac{1}{\text{x}^5}\Big(\frac{\text{x}^4+1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
$=\frac{1}{\text{x}^5}\Big(1+\frac{1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
Let, $\frac{1}{\text{x}^4}=\text{t}$
$\Rightarrow-\frac{4}{\text{x}^5}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{\text{x}^5}\text{ dx}=-\frac{\text{dt}}{4}$
Multiplying and dividing by $x^{-3},$ we obtain
$\frac{\text{x}^{-3}}{\text{x}^2.\text{x}^{-3}(\text{x}^4+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^2.\text{x}^{-3}}$
$=\frac{(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^5.(\text{x}^{4})^{-\frac{3}{4}}}$
$=\frac{1}{\text{x}^5}\Big(\frac{\text{x}^4+1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
$=\frac{1}{\text{x}^5}\Big(1+\frac{1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
Let, $\frac{1}{\text{x}^4}=\text{t}$
$\Rightarrow-\frac{4}{\text{x}^5}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{\text{x}^5}\text{ dx}=-\frac{\text{dt}}{4}$
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