#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 6 Maths Textbook Solution.

Answer: The required value of the integration is,

$\tan ^{-1}\left(\frac{x^{2}-1}{x}\right)+c$

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

Given: The given integral is $I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1}dx$

Solution:

The given equation can be written as,

\begin{aligned} I &=\int \frac{x^{2}+1}{\frac{x^{2}}{\left(\frac{x^{4}-x^{2}+1}{x^{2}}\right)}} d x \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{\left(x^{2}-1+\frac{1}{x^{2}}\right)} \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{x^{2}+\frac{1}{x^{2}}+1-2} \end{aligned}           [divinding $x^{2}$ on both numerator and denominator ]

$=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{\left(x-\frac{1}{x}\right)^{2}+1^{2}}$         [Making the perfect square of $\left ( a+b \right )^{2}$  ]

Now, substitute $x-\frac{1}{x}=t$ then,

$\left (1+\frac{1}{x^{2}} \right )dx=dt$

Now, solving the obtained equation,

\begin{aligned} I &=\int \frac{d t}{t^{2}+1} \\ &=\tan ^{-1} t+c \end{aligned}

Re-substitute $t=x-\frac{1}{x}$ into the above equation.

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

Thus the required value of the given integration is,

$\tan ^{-1}\left(\frac{x^{2}-1}{x}\right)+c$

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