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#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 5 Maths Textbook Solution.

Answer:  The required value of the integral is,

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

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

Given:$I=\int \frac{x^{2}-3 x+1}{x^{4}+x^{2}+1} d x$

Solution: The given equation can be written as,

\begin{aligned} I &=\int \frac{1-\frac{3}{x}+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x-\int \frac{3 x}{x^{4}+x^{2}+1} d x \end{aligned}         [dividing $x^{2}$ i both numerator and denominator]

[ Making the perfect square as $\left ( a+b \right )^{2}$ ]

Now, substituting$x-\frac{1}{x}=t\: and \: x^{2}=z$ then,

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

We knows

$\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$
\begin{aligned} &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C_{1}-\frac{3}{2} \int \frac{d z}{\left(z+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C_{1}-\frac{3}{2} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left|z+\frac{1}{2}\right|}{\frac{\sqrt{3}}{2}}\right)+C_{2} \end{aligned}

\begin{aligned} &t=x-\frac{1}{4} \text { and } z=x^{2} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+C_{1}-\sqrt{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C_{2} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\sqrt{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C \end{aligned}