Question
Evaluate the following integrals:
$\int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}$
$\int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}$
✓
Answer
Let $1+\log\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\text{x}=1\Rightarrow\text{t}=1$
$=\text{x}=2\Rightarrow\text{t}=1+\log2$
$\therefore\ \int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}=\int^\limits{1+\log2}_1\frac{\text{dt}}{\text{t}^2}$
$=\Big[\frac{-1}{\text{t}}\Big]^{1+\log2}_1$
$=\bigg[\frac{-1}{1+\log2}+1\bigg]$
$=\bigg[\frac{-1+1+\log2}{1+\log2}\bigg]$
$=\bigg[\frac{\log2}{1+\log2}\bigg]$ $\big[\because\log\text{e}=1\big]$
$=\frac{\log2}{\log\text{e}+\log2}$ $\big[\log\text{a}+\log\text{b}=\log\text{ab}\big]$
$=\frac{\log2}{\log2\text{e}}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\text{x}=1\Rightarrow\text{t}=1$
$=\text{x}=2\Rightarrow\text{t}=1+\log2$
$\therefore\ \int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}=\int^\limits{1+\log2}_1\frac{\text{dt}}{\text{t}^2}$
$=\Big[\frac{-1}{\text{t}}\Big]^{1+\log2}_1$
$=\bigg[\frac{-1}{1+\log2}+1\bigg]$
$=\bigg[\frac{-1+1+\log2}{1+\log2}\bigg]$
$=\bigg[\frac{\log2}{1+\log2}\bigg]$ $\big[\because\log\text{e}=1\big]$
$=\frac{\log2}{\log\text{e}+\log2}$ $\big[\log\text{a}+\log\text{b}=\log\text{ab}\big]$
$=\frac{\log2}{\log2\text{e}}$
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