If ∫ ((4 e^x + 6 e^(−x) )/(9 e^x − 4 e^(−x) ))dx=Ax+B log(9e^(2x) −4)+C, then


Question and Answers Forum

All Questions Topic List
UNKNOWN Questions
Previous in All Question Next in All Question
Previous in UNKNOWN Next in UNKNOWN
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
$(Ax+Blog(9e^(2x) −4))^′ =((4e^x +6e^(−x) )/(9e^x −4e^(−x) )) but the left hand side is equal to A+B×((18e^(2x) )/(9e^(2x) −4))×(1/(ln10))=A+(B/(ln10))×((18e^x )/(9e^x −4e^(−x) ))= ((9Ae^x ln10−4Ae^(−x) ln10+18Be^x )/(ln10(9e^x −4e^(−x) )))= (((9A+((18)/(ln10))B)e^x −4Ae^(−x) )/(9e^x −4e^(−x) ))⇒ { ((4A=6⇒A=(3/2))),((9A+((18)/(ln10))B=4⇒((18)/(ln10))B=((−19)/2)⇒B=((−19ln10)/(36)))) :}$
${(A x + B l o g (9 e^{2 x} - 4))}^{^{'}} = \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} b u t$ $t h e l e f t h a n d s i d e i s e q u a l t o$ $A + B \times \frac{18 e^{2 x}}{9 e^{2 x} - 4} \times \frac{1}{l n 10} = A + \frac{B}{l n 10} \times \frac{18 e^{x}}{9 e^{x} - 4 e^{- x}} =$ $\frac{9 {A e}^{x} l n 10 - 4 {A e}^{- x} l n 10 + 18 {B e}^{x}}{l n 10 (9 e^{x} - 4 e^{- x})} =$ $\frac{(9 A + \frac{18}{l n 10} B) e^{x} - 4 {A e}^{- x}}{9 e^{x} - 4 e^{- x}} \Rightarrow$ ${\begin{cases} 4 A = 6 \Rightarrow A = \frac{3}{2} \\ 9 A + \frac{18}{l n 10} B = 4 \Rightarrow \frac{18}{l n 10} B = \frac{- 19}{2} \Rightarrow B = \frac{- 19 l n 10}{36} \end{cases}$
Commented by mathmax by abdo last updated on 10/Jul/19

$l e t I = \int \frac{4 e^{x} + 6 e^{- x}}{9 e^{x} - 4 e^{- x}} d x [\Rightarrow I = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} d x c h a n g e m e n t e^{2 x} = t g i v e$ $I = \int \frac{4 t + 6}{9 t - 4} \frac{d t}{2 t} = \int \frac{2 t + 3}{t (9 t - 4)} d t l e t d e c o m p o s e F (t) = \frac{2 t + 3}{t (9 t - 4)}$ $F (t) = \frac{a}{t} + \frac{b}{9 t - 4}$ $a = {l i m}_{t \to 0} t F (t) = - \frac{3}{4}$ $b = {l i m}_{t \to \frac{4}{9}} (9 t - 4) F (t) = \frac{2 \times \frac{4}{9} + 3}{\frac{4}{9}} = \frac{8 + 27}{4} = \frac{35}{4} \Rightarrow$ $F (t) = - \frac{3}{4 t} + \frac{35}{4 (9 t - 4)} \Rightarrow I = \int (- \frac{3}{4 t} + \frac{35}{4 (9 t - 4)}) d t$ $= - \frac{3}{4} l n ∣ t ∣ + \frac{35}{36} l n ∣ 9 t - 4 ∣ + c$ $- \frac{3}{4} (2 x) + \frac{35}{36} l n ∣ 9 e^{2 x} - 4 ∣ + c = - \frac{3}{2} x + \frac{35}{36} l n ∣ 9 e^{2 x} - 4 ∣ + c \Rightarrow$ $A = - \frac{3}{2}, B = \frac{35}{36} a n d C = c .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum