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Question Number 70483 by Mikaell last updated on 04/Oct/19

∫((ln2x)/(ln4x)).(dx/x)

$$\int\frac{{ln}\mathrm{2}{x}}{{ln}\mathrm{4}{x}}.\frac{{dx}}{{x}} \\ $$

Commented by peter frank last updated on 04/Oct/19

thankx

$${thankx} \\ $$

Commented by kaivan.ahmadi last updated on 04/Oct/19

u=lnx⇒du=(dx/x) ln2x=ln2+lnx=u+ln2 ln4x=ln4+lnx=u+ln4 ⇒∫((u+ln2)/(u+ln4))du=∫((udu)/(u+ln4))+∫((ln2du)/(u+ln4))= ∫(1−((ln4)/(u+ln4)))du+ln2∫(du/(u+ln4))= u−ln4ln(u+ln4)+ln2ln(u+ln4)+C= u+(ln2−ln4)ln(u+ln4)+C= u−ln2ln(u+ln4)+C=lnx−ln2ln(lnx+ln4)+C= lnx−ln2ln(ln4x)+C

$${u}={lnx}\Rightarrow{du}=\frac{{dx}}{{x}} \\ $$$${ln}\mathrm{2}{x}={ln}\mathrm{2}+{lnx}={u}+{ln}\mathrm{2} \\ $$$${ln}\mathrm{4}{x}={ln}\mathrm{4}+{lnx}={u}+{ln}\mathrm{4} \\ $$$$\Rightarrow\int\frac{{u}+{ln}\mathrm{2}}{{u}+{ln}\mathrm{4}}{du}=\int\frac{{udu}}{{u}+{ln}\mathrm{4}}+\int\frac{{ln}\mathrm{2}{du}}{{u}+{ln}\mathrm{4}}= \\ $$$$\int\left(\mathrm{1}−\frac{{ln}\mathrm{4}}{{u}+{ln}\mathrm{4}}\right){du}+{ln}\mathrm{2}\int\frac{{du}}{{u}+{ln}\mathrm{4}}= \\ $$$${u}−{ln}\mathrm{4}{ln}\left({u}+{ln}\mathrm{4}\right)+{ln}\mathrm{2}{ln}\left({u}+{ln}\mathrm{4}\right)+{C}= \\ $$$${u}+\left({ln}\mathrm{2}−{ln}\mathrm{4}\right){ln}\left({u}+{ln}\mathrm{4}\right)+{C}= \\ $$$${u}−{ln}\mathrm{2}{ln}\left({u}+{ln}\mathrm{4}\right)+{C}={lnx}−{ln}\mathrm{2}{ln}\left({lnx}+{ln}\mathrm{4}\right)+{C}= \\ $$$${lnx}−{ln}\mathrm{2}{ln}\left({ln}\mathrm{4}{x}\right)+{C} \\ $$$$ \\ $$$$ \\ $$

Commented by Mikaell last updated on 04/Oct/19

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

Answered by Tanmay chaudhury last updated on 04/Oct/19

lnx=t→(dt/dx)=(1/x) ∫((ln2+lnx)/(2ln2+lnx))×(dx/x) ∫((ln2+t)/(2ln2+t))×dt (1/4)∫((2ln2+t+t)/(2ln2+t))dt (1/4)∫dt+(1/4)∫((2ln2+t−2ln2)/(2ln2+t))dt (1/2)∫dt−((ln2)/2)∫(dt/(2ln2+t)) (t/2)−((ln2)/2)ln(2ln2+t)+C ((lnx)/2)−((ln2)/2)ln(2ln2+lnx)+c

$${lnx}={t}\rightarrow\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{x}} \\ $$$$\int\frac{{ln}\mathrm{2}+{lnx}}{\mathrm{2}{ln}\mathrm{2}+{lnx}}×\frac{{dx}}{{x}} \\ $$$$\int\frac{{ln}\mathrm{2}+{t}}{\mathrm{2}{ln}\mathrm{2}+{t}}×{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{2}{ln}\mathrm{2}+{t}+{t}}{\mathrm{2}{ln}\mathrm{2}+{t}}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{dt}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{2}{ln}\mathrm{2}+{t}−\mathrm{2}{ln}\mathrm{2}}{\mathrm{2}{ln}\mathrm{2}+{t}}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{dt}−\frac{{ln}\mathrm{2}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{2}{ln}\mathrm{2}+{t}} \\ $$$$\frac{{t}}{\mathrm{2}}−\frac{{ln}\mathrm{2}}{\mathrm{2}}{ln}\left(\mathrm{2}{ln}\mathrm{2}+{t}\right)+{C} \\ $$$$\frac{{lnx}}{\mathrm{2}}−\frac{{ln}\mathrm{2}}{\mathrm{2}}{ln}\left(\mathrm{2}{ln}\mathrm{2}+{lnx}\right)+{c} \\ $$

Commented by peter frank last updated on 04/Oct/19

thanx

$${thanx} \\ $$

Commented by Mikaell last updated on 04/Oct/19

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

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