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Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 7 Maths Textbook Solution.

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Answer: The required value of the integration is,

\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c

Hint: Use the identity  formula \int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+c


            I=\int \frac{x^{2}-1}{x^{4}+1} d x


Solution: Given equation can be written as

I=\frac{x^{2}-1}{x^{4}+1} d x

Divide numerator and denominator by x^{2}

\begin{aligned} I &=\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} d x \\ &=\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+2-2} d x \\ &=\int \frac{\left(1-\frac{1}{x^{2}}\right)}{\left(x+\frac{1}{x}\right)^{2}-2} d x \end{aligned}


\left ( 1-\frac{1}{x^{2}}\right )dx=dt     differentiating on both sides with respect to t,

\begin{aligned} I &=\int \frac{d t}{t^{2}-2} \\ &=\int \frac{d t}{(t)^{2}-\left(\sqrt{2)^{2}}\right.} \\ &=\frac{1}{2 \sqrt{2}} \log \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+c \end{aligned}

Re- substitutet=x+\frac{1}{x}  into the above equation.

\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c

Where c is an integrating constant.

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