11what i did was
we have
9e2x+5/9e2x-1
we add and subtract 6 4m the above equation
∫9e2x-1+6/9e2x-1
∫dx + ∫6/9e2x-1
x+6ln(9e2x-1)
hence getting B =6
please find my mistake (iff exists)
∫[(9ex+5e-x)/(9ex-e-x)]dx=Ax+B ln(9e2x-1)+C
Then which of the following is true
-
UP 0 DOWN 0 0 9
9 Answers
1B=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
1u see multiply both the numerator and denominator by ex
u get
(9e2x+5)/(9e2x-1) = 1+(6/(9e2x-1))
now in second term divide both numerator and denominator by e2x and put 9-e-2x=t.. .
you willl get the reqd form
11
11How is that true?
is ∫dx/9e2x-1 =ln(9e2x-1)
shouldnt there be a 18e2x term in the numerator
9u neednt integrate rohan
differentiate RHS and compare ull get A and B