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

∫((ln2x)/(ln4x)).(dx/x)

Commented bypeter frank last updated on 04/Oct/19

thankx

Commented bykaivan.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

Commented byMikaell last updated on 04/Oct/19

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

Commented bypeter frank last updated on 04/Oct/19

thanx

Commented byMikaell last updated on 04/Oct/19

thank you Sir