Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 105574 by bemath last updated on 30/Jul/20

∫ ((x−1)/(x+x^2 ln x)) dx ?

$$\int\:\frac{{x}−\mathrm{1}}{{x}+{x}^{\mathrm{2}} \mathrm{ln}\:{x}}\:{dx}\:?\: \\ $$

Answered by bobhans last updated on 30/Jul/20

set x =e^u , dx = e^u du  I=∫ (((e^u −1)e^u )/(e^u +ue^(2u) )) du = ∫ ((e^u −1)/(1+ue^u )) du  I=∫(((e^u +ue^u )/(1+ue^u ))−1) du  substitute v = 1+ue^u   I=∫ (dv/v)−∫du = ln v−u + c  I= ln (1+ue^u )−u+c  I= ln (1+xln x)−ln x + c ★  I=ln ((1/x)+ln x)+c

$${set}\:{x}\:={e}^{{u}} ,\:{dx}\:=\:{e}^{{u}} {du} \\ $$$${I}=\int\:\frac{\left({e}^{{u}} −\mathrm{1}\right){e}^{{u}} }{{e}^{{u}} +{ue}^{\mathrm{2}{u}} }\:{du}\:=\:\int\:\frac{{e}^{{u}} −\mathrm{1}}{\mathrm{1}+{ue}^{{u}} }\:{du} \\ $$$${I}=\int\left(\frac{{e}^{{u}} +{ue}^{{u}} }{\mathrm{1}+{ue}^{{u}} }−\mathrm{1}\right)\:{du} \\ $$$${substitute}\:{v}\:=\:\mathrm{1}+{ue}^{{u}} \\ $$$${I}=\int\:\frac{{dv}}{{v}}−\int{du}\:=\:\mathrm{ln}\:{v}−{u}\:+\:{c} \\ $$$${I}=\:\mathrm{ln}\:\left(\mathrm{1}+{ue}^{{u}} \right)−{u}+{c} \\ $$$${I}=\:\mathrm{ln}\:\left(\mathrm{1}+{x}\mathrm{ln}\:{x}\right)−\mathrm{ln}\:{x}\:+\:{c}\:\bigstar \\ $$$${I}=\mathrm{ln}\:\left(\frac{\mathrm{1}}{{x}}+\mathrm{ln}\:{x}\right)+{c}\: \\ $$

Terms of Service

Privacy Policy