Test:351 QNo:67

B=2 aaya tha rite hai

rhs ko diff kar n then match
u can rite it as

int.... (1)dx +(9e^x+6e^-x)dx/(9e^x-e^-x)

nw difftntaiting d rhs
u get

A + B(18e^x)/(9e^2x-1)

so 18b=9 aata hai

so b=2

u see multiply both the numerator and denominator by e^x

u get

(9e^2x+5)/(9e^2x-1) = 1+(6/(9e^2x-1))

now in second term divide both numerator and denominator by e^2x and put 9-e^-2x=t.. .

you willl get the reqd form

How is that true?

is ∫dx/9e^2x-1 =ln(9e^2x-1)

shouldnt there be a 18e^2x term in the numerator

c post 2#

ho ya ~~~~~~~~~~~~ u ve already said that :)

sry dint see

i know that celestine ..

just gave the method ..