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$\int(\operatorname{In} x)^{-1} \mathrm{~d} x-\int(\operatorname{In} x)^{-2} \mathrm{~d} x$ is equal to
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Verified Answer
The correct answer is:
$x(\operatorname{In} x)^{-1}+c$
$\int(\ln x)^{-1} \mathrm{dx}-\int(\ln x)^{-2} \mathrm{dx}$
$=\int\left[\frac{1}{\ell^{n} x}-\frac{1}{(\ell n x)^{2}}\right] \cdot d x$
Put $\ell \mathrm{n} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{x}=\mathrm{e}^{\mathrm{t}}$
$\mathrm{dx}=\mathrm{e}^{\mathrm{l}} \cdot \mathrm{dt}$
$\therefore \int\left[\frac{1}{\ln x}-\frac{1}{(\ell n x)^{2}}\right] \mathrm{dx}=\int\left(\frac{1}{t}-\frac{1}{t^{2}}\right) \cdot \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt}$
$=\int \mathrm{e}^{\mathrm{t}} \cdot\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}^{2}}\right) \mathrm{dt}$
$=\frac{\mathrm{e}^{\mathrm{t}}}{\mathrm{t}}+\mathrm{c}=\frac{\mathrm{x}}{\ln \mathrm{x}}+\mathrm{c}$
$=x(\ln x)^{-1}+c$
$=\int\left[\frac{1}{\ell^{n} x}-\frac{1}{(\ell n x)^{2}}\right] \cdot d x$
Put $\ell \mathrm{n} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{x}=\mathrm{e}^{\mathrm{t}}$
$\mathrm{dx}=\mathrm{e}^{\mathrm{l}} \cdot \mathrm{dt}$
$\therefore \int\left[\frac{1}{\ln x}-\frac{1}{(\ell n x)^{2}}\right] \mathrm{dx}=\int\left(\frac{1}{t}-\frac{1}{t^{2}}\right) \cdot \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt}$
$=\int \mathrm{e}^{\mathrm{t}} \cdot\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}^{2}}\right) \mathrm{dt}$
$=\frac{\mathrm{e}^{\mathrm{t}}}{\mathrm{t}}+\mathrm{c}=\frac{\mathrm{x}}{\ln \mathrm{x}}+\mathrm{c}$
$=x(\ln x)^{-1}+c$
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