Question

integration of dx /x√x^4-1=​

Answer 1

$\underline{\textsf{Solution :}}$

$\textsf{Let us take}\:\boxed{\mathsf{x^{2}= sec\theta}}$

$\textsf{Taking differentials, we get}$

$\mathsf{2x\:dx = sec\theta\:tan\theta\:d\theta}$

$\to \mathsf{dx=\frac{sec\theta\:tan\theta\:d\theta}{2x}}$

$\to \boxed{\mathsf{dx = \frac{sec\theta\:tan\theta\:d\theta}{2\sqrt{sec\theta}}}}$

$\mathsf{Now,\:\int \frac{dx}{x\sqrt{x^{4}-1}}}$

$\mathsf{=\int \frac{\frac{sec\theta\:tan\theta\:d\theta}{2\sqrt{sec\theta}}}{\sqrt{sec\theta}\sqrt{sec^{2}\theta-1}}}$

$\mathsf{=\frac{1}{2} \int \frac{sec\theta\:tan\theta\:d\theta}{\sqrt{sec\theta}\sqrt{sec\theta}\sqrt{tan^{2}\theta}}}$

$\mathsf{=\frac{1}{2} \int \frac{sec\theta\:tan\theta\:d\theta}{sec\theta\:tan\theta}}$

$\mathsf{=\frac{1}{2} \int d\theta}$

$\mathsf{=\frac{1}{2} \theta + C}$

$\textsf{where C is integral constant}$

$\mathsf{=\frac{1}{2} sec^{-1}(x^{2})+C}$

$\to \boxed{\mathsf{\int \frac{dx}{x\sqrt{x^{4}-1}}=\frac{1}{2} sec^{-1}(x^{2})+C}}$

$\textsf{which is the required integral.}$

Answer 2

Answer:$\frac{1}{2}\sec^{-1}(x^2)+c$

Step-by-step explanation:

Let,

$I=\int{\frac{dx}{x\sqrt{x^4-1}}}$

Dividing  denominator and  numerator by x,

$I=\int{\frac{x\;dx}{x^2\sqrt{x^4-1}}}$

Putting x² = t ;

2x = dt/dx

x.dx = dt/2

Now,

$I=\frac{1}{2}\int{\frac{dt}{t\sqrt{t^2-1}}}\\\;\\I=\frac{1}{2}.\sec^{-1}(t)+c\\\;\\I=\frac{1}{2}\sec^{-1}(x^2)+c$

Note:- $\int{\frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}(x)$