Question
Evaluate the following integrals:
$\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
$\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
✓
Answer
Let $\text{I}=\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$ Then,
Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$
$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$
$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
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