Question Number 93481 by mashallah last updated on 13/May/20
$$\int\left(\mathrm{log}\:\mathrm{x}/\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}= \\ $$
Commented by abdomathmax last updated on 15/May/20
$${I}\:=\int\:\frac{{lnx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${I}\:=−\frac{{lnx}}{{x}}\:−\int\:\left(−\frac{\mathrm{1}}{{x}}\right)×\frac{{dx}}{{x}}\:=−\frac{{lnx}}{{x}}\:+\int\:\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$$=−\frac{{lnx}}{{x}}−\frac{\mathrm{1}}{{x}}\:+{C}\:=−\frac{\mathrm{1}+{lnx}}{{x}}\:+{C} \\ $$
Answered by john santu last updated on 13/May/20
![∫ ((ln x)/x^2 ) dx = ∫ ln(x).x^(−2) dx =W [ by parts ] u = ln(x) ⇒du = (dx/x) v= ∫ x^(−2) dx = −x^(−1) W= −x^(−1) ln(x) + ∫ x^(−1) . (dx/x) W= −((ln(x))/x) −(1/x) + c W = − ((ln(x)+1)/x) + c](Q93508.png)
$$\int\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int\:\mathrm{ln}\left(\mathrm{x}\right).\mathrm{x}^{−\mathrm{2}} \:\mathrm{dx}\:=\mathrm{W} \\ $$$$\left[\:\mathrm{by}\:\mathrm{parts}\:\right]\: \\ $$$$\mathrm{u}\:=\:\mathrm{ln}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{du}\:=\:\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\mathrm{v}=\:\int\:\mathrm{x}^{−\mathrm{2}} \:\mathrm{dx}\:=\:−\mathrm{x}^{−\mathrm{1}} \\ $$$$\mathrm{W}=\:−\mathrm{x}^{−\mathrm{1}} \:\mathrm{ln}\left(\mathrm{x}\right)\:+\:\int\:\mathrm{x}^{−\mathrm{1}} \:.\:\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\mathrm{W}=\:−\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}}\:−\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{c}\: \\ $$$$\mathrm{W}\:=\:−\:\frac{\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{c}\: \\ $$