If-4-e-x-6-e-x-9-e-x-4-e-x-dx-Ax-B-log-9e-2x-4-C-then-

Question Number 63865 by mmkkmm000m last updated on 10/Jul/19

If \int \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} d x = A x + B \log (9 e^{2 x} - 4) + C,

then

Commented by kaivan.ahmadi last updated on 10/Jul/19

{(A x + B l o g (9 e^{2 x} - 4))}^{^{'}} = \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} b u t

t h e l e f t h a n d s i d e i s e q u a l t o

A + B \times \frac{18 e^{2 x}}{9 e^{2 x} - 4} \times \frac{1}{l n 10} = A + \frac{B}{l n 10} \times \frac{18 e^{x}}{9 e^{x} - 4 e^{- x}} =

\frac{9 {A e}^{x} l n 10 - 4 {A e}^{- x} l n 10 + 18 {B e}^{x}}{l n 10 (9 e^{x} - 4 e^{- x})} =

\frac{(9 A + \frac{18}{l n 10} B) e^{x} - 4 {A e}^{- x}}{9 e^{x} - 4 e^{- x}} \Rightarrow

{\begin{cases} 4 A = 6 \Rightarrow A = \frac{3}{2} \\ 9 A + \frac{18}{l n 10} B = 4 \Rightarrow \frac{18}{l n 10} B = \frac{- 19}{2} \Rightarrow B = \frac{- 19 l n 10}{36} \end{cases}

Commented by mathmax by abdo last updated on 10/Jul/19

l e t I = \int \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} d x [\Rightarrow I = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} d x c h a n g e m e n t e^{2 x} = t g i v e

I = \int \frac{4 t + 6}{9 t - 4} \frac{d t}{2 t} = \int \frac{2 t + 3}{t (9 t - 4)} d t l e t d e c o m p o s e F (t) = \frac{2 t + 3}{t (9 t - 4)}

F (t) = \frac{a}{t} + \frac{b}{9 t - 4}

a = {l i m}_{t \to 0} t F (t) = - \frac{3}{4}

b = {l i m}_{t \to \frac{4}{9}} (9 t - 4) F (t) = \frac{2 \times \frac{4}{9} + 3}{\frac{4}{9}} = \frac{8 + 27}{4} = \frac{35}{4} \Rightarrow

F (t) = - \frac{3}{4 t} + \frac{35}{4 (9 t - 4)} \Rightarrow I = \int (- \frac{3}{4 t} + \frac{35}{4 (9 t - 4)}) d t

= - \frac{3}{4} l n ∣ t ∣ + \frac{35}{36} l n ∣ 9 t - 4 ∣ + c

- \frac{3}{4} (2 x) + \frac{35}{36} l n ∣ 9 e^{2 x} - 4 ∣ + c = - \frac{3}{2} x + \frac{35}{36} l n ∣ 9 e^{2 x} - 4 ∣ + c \Rightarrow

A = - \frac{3}{2}, B = \frac{35}{36} a n d C = c .