#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 1 Maths Textbook Solution.

Answer: The required value of the integral is,

$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c$

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

Given:

$\int \frac{\left ( x^{2}+1 \right )}{\left ( x^{4}+x^{2}+1 \right )}dx$

Solution: The equation can be written as

$I=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$      [ dividing $x^{2}$ both denominator and denominator]

$=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x^{2}}\right)^{2}+3} d x$      [Making the perfect square as $(a+b)^{2}$ ]

Let $x-\frac{1}{x}=t$

Now differentiating both side w.r.t t

\begin{aligned} \left(1+\frac{1}{x^{2}}\right) &=d t \\ I &=\int \frac{1}{t^{2}+3} d t \end{aligned}

On using standard identity we get, $I=\frac{1}{\sqrt{3}}\left ( \frac{t}{\sqrt{3}} \right )+c$

Substituting $t=x-\frac{1}{x}$ we get,

\begin{aligned} &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\left(x-\frac{1}{x}\right)}{\sqrt{3}}\right)+c \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c \end{aligned}

So, the required value of the integration is,

$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c$