Question
Evaluate the following integrals:$\int\frac{\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}}{\text{x}^2+\text{x}+3}\text{ dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{x}+3\big)+\mu$
$\text{x}+1=\lambda(2\text{x}+1)+\mu$
$\text{x}+1=(2\lambda)\text{x}+(\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=1$
$\Rightarrow\lambda=\frac{1}{2}$
$\lambda+\mu=1$
$\Rightarrow\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x}+1)+\frac{1}{2}}{\text{x}^2+\text{x}+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{11}{4}\big)}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{\sqrt{11}}{2}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{2}\times\frac{1}{\Big(\frac{\sqrt{11}}{2}\Big)}\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{\sqrt{11}}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt{11}}\Big)+\text{C}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{x}+3\big)+\mu$
$\text{x}+1=\lambda(2\text{x}+1)+\mu$
$\text{x}+1=(2\lambda)\text{x}+(\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=1$
$\Rightarrow\lambda=\frac{1}{2}$
$\lambda+\mu=1$
$\Rightarrow\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x}+1)+\frac{1}{2}}{\text{x}^2+\text{x}+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{11}{4}\big)}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{\sqrt{11}}{2}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{2}\times\frac{1}{\Big(\frac{\sqrt{11}}{2}\Big)}\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{\sqrt{11}}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt{11}}\Big)+\text{C}$
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