A/(b+ce^x) dx = ? Please solve this

1/ab en (c/be^-x+c)

Step-by-step explanation:

Sure is a harsh question! Here's my Explanation

b+ce^x = t

ce^x an = dt

e^xan = dt/c

an = dt/ce^x = dt/c(t-b/c) = at/(t-b)

en = t-b/c

A/b+ce^x dx = a/t dt/t-b

a ∫1/t (t-b) dt = 1/a∫ (1/(t-b) - 1/t) dt

= 1/ab [∫1/(t-b) dt + ∫-1/t dt]

= 1/ab [en (t-b) - en(t)]

= 1/ab en ((t-b)/t)

t = b + ce^x

= 1/ab en (b+ce^x -b/b+ce^x)

=1/ab en (ce^x/b+ce^x)

= 1/ab en (c/be^-x+c)

w'x is 0.

hope it

i think it is 18

step-by-step explanation: