#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 8 Maths Textbook Solution.

Answer: $\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$

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

Hint: Useing $\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$

Explanation:

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

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

$\therefore \int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-2 x+1} d x=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-\frac{2}{3}}{x^{2}-x+1} d x$

$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1+1-\frac{2}{3}}{x^{2}-x+1} d x$

$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{3}{2} \times \frac{1}{3} \int \frac{1}{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x$

$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$

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

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

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