Question
Evaluate the following definite integrals:
$\int_{1}^\limits{3}\frac{\log\text{x}}{(\text{x}+1)^2}\text{ dx}$
$\int_{1}^\limits{3}\frac{\log\text{x}}{(\text{x}+1)^2}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{1}^\limits{3}\frac{\log\text{x}}{(\text{x}+1)^2}\text{ dx}$ Then,
$\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1-\int_{1}^\limits{3}\Big(\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\frac{1}{\text{x}(\text{x}+1)}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\big[\log\text{x}-\log(\text{x}+1)\big]^3_1$
$\Rightarrow\text{I}=\frac{-1}{4}\log3+\log3-\log4+\log2$
$\Rightarrow\text{I}=\frac{3}{4}\log3-\log2$
$\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1-\int_{1}^\limits{3}\Big(\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\frac{1}{\text{x}(\text{x}+1)}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\big[\log\text{x}-\log(\text{x}+1)\big]^3_1$
$\Rightarrow\text{I}=\frac{-1}{4}\log3+\log3-\log4+\log2$
$\Rightarrow\text{I}=\frac{3}{4}\log3-\log2$
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