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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 65 maths textbook solution

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

Hint: Use substitution method to solve this integral

Given: \int \frac{1}{x \sqrt{x^{4}-1}} d x


        \begin{aligned} &\operatorname{let} I=\int \frac{1}{x \sqrt{x^{4}-1}} d x=\int \frac{1}{x \sqrt{\left(x^{2}\right)^{2}-1}} d x \\ &\text { Putting } x^{2}=t \Rightarrow 2 x \; d x=d t \Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}

        I=\int \frac{1}{x \sqrt{t^{2}-1}} \frac{d t}{2 x}=\frac{1}{2} \int \frac{1}{x^{2} \sqrt{t^{2}-1}} d t

            =\frac{1}{2} \int \frac{1}{t \sqrt{t^{2}-1}} d t \quad\left[\because x^{2}=t\right]

            =\frac{1}{2} \sec ^{-1} \mathrm{t}+c \quad\left[\because \int \frac{1}{x \sqrt{x^{2}-1}} d x=\sec ^{-1} x+c\right]

            =\frac{1}{2} \sec ^{-1}\left(x^{2}\right)+c \quad\left[\because t=x^{2}\right]

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