Evaluate the following definite integrals: _ 1 ^ 2 x+3 x(x+2) dx - Vidyadip

Question

Evaluate the following definite integrals:
$\int\limits_{1}^{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{ dx}$

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Answer

We have
$\int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}$
$=\int_{1}^\limits{2}\frac{\text{x}}{\text{x}(\text{x}+2)}\text{ dx}+\int_{1}^\limits{2}\frac{3}{\text{x}(\text{x}+2)}\text{ dx}$
$=\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+2)}\text{ dx}+\int_{1}^\limits{2}\frac{3}{\text{x}(\text{x}+2)}\text{ dx}$
$=\big[\log(\text{x}+2)\big]^2_1+\frac{3}{2}\int_{1}^\limits{2}\frac{1}{\text{x}}-\frac{1}{\text{x}+2}\text{ dx}$ [using partial fraction]
$=\big[\log(\text{x}+2)\big]^2_1+\Big[\frac{3}{2}\log\text{x}-\frac{3}{2}\log(\text{x}+2)\Big]^2_1$
$=\Big[\frac{3}{2}\log\text{x}-\frac{1}{2}\log(\text{x}+2)\Big]^2_1$
$=\frac{1}{2}\big[3\log2-\log4+\log3\big]$
$=\frac{1}{2}\big[3\log2-2\log2+\log3\big]$ $\big[\because\log4=2\log2\big]$
$=\frac{1}{2}\big[\log2+\log3\big]$
$=\frac{1}{2}\big[\log6\big]$
$=\frac{1}{2}\log6$
$\therefore\ \int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}=\frac{1}{2}\log6$

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