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
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