Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$ $=\int\begin{bmatrix}1+\frac{\text{2x}}{\text{x}^2-\text{x}+1}\end{bmatrix}\text{dx}$ $\text{I}=\text{x}+\int\frac{2\text{x}}{\text{x}^2-\text{x}+1}\text{ dx}+\text{C}_1\ ....(1)$ Let $\text{I}=\int\frac{2\text{x}}{\text{x}^2-\text{x}+1}\text{ dx}$ Let $2\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}+1\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $2\text{x}=(2\lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$2=2\lambda$ $\Rightarrow\lambda=1$ $-\lambda+\mu=0$ $\Rightarrow-1+\mu=0$ $\mu=1$ So, $\text{I}_1=\int\frac{(2\text{x}-1)+1}{\text{x}^2-\text{x}+1}\text{ dx}$ $=\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}+1}\text{ dx}+\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}\text{ dx}$ $\text{I}_1=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx+}\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\Big(\frac{\sqrt3}{2}\Big)^2}\text{ dx}$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt3}{2}}\bigg)+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}_2\ .....(2)$ Using equation (1) and (2) $\text{I}=\text{x}+\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}$
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