Evaluate the following integrals: 1 x x^4-1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}$

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Answer

$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}\ ....(1)$
Let $\text{x}^2=\text{t}$ then,
$\text{d}\big(\text{x}^2\big)=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2\text{x}}$
Putting $\text{x}^2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{2\text{x}}$ in equation (1), we get,
$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{t}^2-1}}\times\frac{\text{dt}}{2\text{x}}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\int\frac{1}{\text{t}\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\sec^{-1}\text{t}+\text{C}$
$=\frac{1}{2}\sec^{-1}\text{x}^2+\text{C}$
$\text{I}=\frac{1}{2}\sec^{-1}\big(\text{x}^2\big)+\text{C}$

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