Evaluate the following integrals: x^3+x^2+2x+1 x^2-x+1 dx

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