Question-172635

Question Number 172635 by Mikenice last updated on 29/Jun/22

Answered by MJS_new last updated on 29/Jun/22

\int \frac{4^{x}}{4^{x} + 1} d x =

[t = 4^{x} + 1 \to d x = \frac{d t}{4^{x} \ln 4}]

= \frac{1}{\ln 4} \int \frac{d t}{t} = \frac{\ln t}{\ln 4} =

= \frac{\ln (4^{x} + 1)}{\ln 4} + C

Answered by floor(10²Eta[1]) last updated on 29/Jun/22

let u = 4^{x} + 1 \Rightarrow du = 4^{x} \ln 4 dx

\int \frac{4^{x}}{4^{x} + 1} dx = \frac{1}{\ln 4} \int \frac{du}{u} = \frac{\ln ∣ u ∣}{\ln 4} = \frac{\ln (4^{x} + 1)}{\ln 4} = \log_{4} (4^{x} + 1) + C

Answered by Mathspace last updated on 30/Jun/22

I = \int \frac{4^{x}}{1 + 4^{x}} d x w e d o t h e c h a n g e m e n t

4^{x} = t \Rightarrow e^{x l n (4)} = t \Rightarrow x l n 4 = l n t \Rightarrow

x = \frac{l n t}{2 l n (2)} \Rightarrow

I = \int \frac{t}{t + 1} \frac{d t}{2 t l n 2}

= \frac{1}{2 l n 2} \int \frac{d t}{t + 1} = \frac{1}{2 l n 2} l n ∣ t + 1 ∣ + c

= \frac{1}{2 l n 2} l n (1 + 4^{x}) + c

Commented by CElcedricjunior last updated on 30/Jun/22

o u

= x - \frac{1}{\ln 2} \ln ∣ \frac{2^{x}}{\sqrt{4^{x} + 1}} ∣ + c

\dots \dots l e c \overset{´}{e l} \overset{`}{e b r e} cedric junior \dots \dots \dots .

Answered by CElcedricjunior last updated on 30/Jun/22

\int \frac{4^{x}}{4^{x} + 1} d x = \frac{1}{l n 4} \int \frac{l n 4 e^{x l n 4} d x}{e^{x l n 4} + 1}

= \frac{1}{\ln 4} \ln ∣ 4^{x} + 1 ∣ + C a v e c C \in R

L e c \overset{´}{e} l \overset{`}{e b r e} c e d r i c j u n i o r