Evaluate the following integrals: x+2 2x^2+6x+5 dx - Vidyadip

Question

Evaluate the following integrals:$\int\frac{\text{x}+2}{2\text{x}^2+6\text{x}+5}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}+2}{2\text{x}^2+6\text{x}+5}\text{ dx}$
Let $\text{x}+2=\lambda\frac{\text{d}}{\text{dx}}\big(2\text{x}^2+6\text{x}+5\big)+\mu$
$=\lambda(4\text{x}+6)+\mu$
$\text{x}+2=(4\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$4\lambda=1\Rightarrow\lambda=\frac{1}{4}$
$6\lambda+\mu=2\Rightarrow6\Big(\frac{1}{4}\Big)+\mu=2$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{\frac{1}{4}(4\text{x}+6)+\frac{1}{2}}{2\text{x}^2+6\text{x}+5}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{2}\int\frac{1}{2\text{x}^2+6\text{x}+5}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+3\text{x}+\frac{5}{2}}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{3}{2}\big)+\big(\frac{3}{2}\big)^2-\big(\frac{3}{2}\big)^3+\frac{5}{2}}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\big(\text{x}+\frac{3}{2}\big)^2+\frac{1}{4}}\text{ dx}$
$\text{I}=\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\big(\text{x}+\frac{3}{2}\big)^2+\big(\frac{1}{2}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\times\frac{1}{\frac{1}{2}}\tan^{-1}\bigg(\frac{\text{x}+\frac{3}{2}}{\frac{1}{2}}\bigg)+\text{C dx}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\big|2\text{x}^2+6\text{x}+5\big|+\frac{1}{2}\tan^{-1}(2\text{x}+3)+\text{C}$

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