Menu Close

Question-67069




Question Number 67069 by mRDv143 last updated on 22/Aug/19
Commented by Prithwish sen last updated on 22/Aug/19
∫(((1+logx)^2 )/(1+xlogx+logx+x(logx)^2 )) dx  = ∫(((1+logx)^2 )/((1+logx)(1+xlogx))) dx  =∫((1+logx)/(1+xlogx)) dx  put 1+xlogx = u⇒(1+logx)dx= du  =∫(du/u) = ln∣1+logx∣+C   please check.
$$\int\frac{\left(\mathrm{1}+\mathrm{logx}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{xlogx}+\mathrm{logx}+\mathrm{x}\left(\mathrm{logx}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$=\:\int\frac{\left(\mathrm{1}+\mathrm{logx}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{logx}\right)\left(\mathrm{1}+\mathrm{xlogx}\right)}\:\mathrm{dx} \\ $$$$=\int\frac{\mathrm{1}+\mathrm{logx}}{\mathrm{1}+\mathrm{xlogx}}\:\mathrm{dx}\:\:\mathrm{put}\:\mathrm{1}+\mathrm{xlogx}\:=\:\mathrm{u}\Rightarrow\left(\mathrm{1}+\mathrm{logx}\right)\mathrm{dx}=\:\mathrm{du} \\ $$$$=\int\frac{\mathrm{du}}{\mathrm{u}}\:=\:\mathrm{ln}\mid\mathrm{1}+\mathrm{logx}\mid+\mathrm{C}\:\:\:\mathrm{please}\:\mathrm{check}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *