Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { 1 }{ x\sqrt { { x }^{ 4 }-1 } } } dx$

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{x\sqrt{{x}^{4}-1}}}$

Let $\sec{t}={x}^{2}\Rightarrow\,\sec{t}\tan{t}dt=2x\,dx$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{\dfrac{\sec{t}\tan{t}}{x}}{x\sqrt{{\sec}^{2}{t}-1}}dt}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{\sec{t}\tan{t}dt}{{x}^{2}\sqrt{{\sec}^{2}{t}-1}}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{\sec{t}\tan{t}dt}{\sec{t}\sqrt{{\sec}^{2}{t}-1}}}$ where ${x}^{2}=\sec{t}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{\tan{t}dt}{\sqrt{{\tan}^{2}{t}}}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{\tan{t}dt}{\tan{t}}}$

$=\dfrac{1}{2}\displaystyle\int{dt}$

$=\dfrac{1}{2}t+c$ where ${x}^{2}=\sec{t}$

$=\dfrac{1}{2}{\sec}^{-1}{\left({x}^{2}\right)}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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