Question

1/6e²ˣ+5eˣ+1 ,Integrate the given function defined over a proper domain w.r.t. x.

Answer 1

HELLO DEAR,




your questions is-----------eˣ.dx/6e²ˣ+5eˣ+1


GIVEN:-
∫1/6e²ˣ+5eˣ+1

put e^x = t
e^x.dx = dt

therefore, I = dt/(6t² + 5t + 1)

=> I = dt/(6t² + 3t + 2t + 1)

=> I = dt/(3t + 1)(2t + 1)

by partial fraction;

1/(3t + 1)(2t + 1) = A/(3t + 1) + B/(2t + 1)

1 = A(2t + 1) + B(3t + 1)

1 = t(2A + 3B) + (A + B)

on comparing both side like power of cofficient of t,

2A + 3B = 0 , A + B = 1

2(1 - B) + 3B = 0 , A = -B

2 - 2B + 3B = 0 , A = -B

B = -2 , A = 2,


therefore, I = ∫2.dt/(3t + 1) - ∫2dt/(2t + 1)

=> I = 2/3log|3t + 1| - 2/2log|2t + 1| + C.

put the value of t = e^x

=> I = 2/3log|3e^x + 1| - log|2e^x + 1| + C.


I HOPE ITS HELP YOU DEAR,
THANKS