Question
Evaluate the following integrals:
$\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}$
$\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}$
✓
Answer
Let I $=\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}\ .....(1)$
Let $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ then,
$\text{d}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]=\text{dt}$
$\Rightarrow\frac{1}{1+\frac{1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{1}{\frac{\text{x}+1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{-\text{x}}{\text{x}^2(\text{x}+1)}\text{dx}=-\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$
Putting $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ and $\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\text{tx}-\text{dt}$
$=-\frac{\text{t}^2}{2}+\text{C}$
$=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
$\therefore\text{I}=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
Let $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ then,
$\text{d}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]=\text{dt}$
$\Rightarrow\frac{1}{1+\frac{1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{1}{\frac{\text{x}+1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{-\text{x}}{\text{x}^2(\text{x}+1)}\text{dx}=-\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$
Putting $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ and $\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\text{tx}-\text{dt}$
$=-\frac{\text{t}^2}{2}+\text{C}$
$=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
$\therefore\text{I}=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
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