Question

Q1.integration of dx /x√x^4-1=​

Answer 1

Step-by-step explanation:

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,

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

Note:-

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

Answer 2

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