Question:
Evaluate the following integrals:
$\int \frac{x+1}{x(x+\log x)} d x$
Solution:
Assume $x+\log x=t$
$\mathrm{d}(x+\log x)=\mathrm{dt}$
$\Rightarrow\left(1+\frac{1}{x}\right) d x=d t$
$\Rightarrow\left(\frac{x+1}{x}\right) d x=d t$
Put $t$ and $d t$ in the given equation we get
$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$
$=\ln |\mathrm{t}|+\mathrm{c}$
But $t=x+\log x$
$=\ln |x+\log x|+c$
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