Question
Evaluate the following integrals:

$\int\frac{\text{x}^3+\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}+1}\text{ dx}$

Answer

Let $\text{I}=\int\frac{\text{x}^3+\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}+1}\text{ dx}$

$=\int\Big[\text{x}+2+\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\Big]\text{dx}$

$\text{I}=\frac{\text{x}^2}{2}+2\text{x}+\int\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}+\text{C}_1\ ....(1)$

Let $\text{I}_1=\int\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}$

Let $3\text{x}-1=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}+1\big)+\mu$

$=\lambda(2\text{x}-1)+\mu$

$3\text{x}-1=(2\lambda)\text{x}-\lambda+\mu$

Comparing the coefficients of like powers of x,

$3=2\lambda\Rightarrow\lambda=\frac{3}{2}$

$-\lambda+\mu=-1\Rightarrow-\Big(\frac{3}{2}\Big)+\mu=-1$

$\mu=\frac{1}{2}$

So, $\text{I}_1=\int\frac{\frac{3}{2}(2\text{x}-1)+\frac{1}{2}}{\text{x}^2-\text{x}+1}\text{ dx}$

$\text{I}_1=\frac{3}{2}\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}+1}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}\text{ dx}$

$\text{I}_1=\frac{3}{2}\int\frac{2\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\Big(\frac{\sqrt3}{2}\Big)^2}\text{ dx}$

$\text{I}_1=\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{2}\times\frac{2}{\sqrt3}\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt3}{2}}\bigg)+\text{C}_2$

$\Big[\text{since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$

$\text{I}_1=\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt3}\Big)+\text{C}_2\ .....(2)$

Using equation (1) and (2)

$\text{I}=\frac{\text{x}^2}{2}+2\text{x}+\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt3}\Big)+\text{C}$

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