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Solve the logarithmic equation $y=\ln\left(\frac{e^{4x}-1}{e^{4x}+1}\right)$

Step-by-step Solution

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Final answer to the problem

$\left(e^{4x}-1\right)\ln\left(e^{4x}-1\right)+\left(-e^{4x}-1\right)\ln\left(e^{4x}+1\right)+C_1$
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Step-by-step Solution

How should I solve this problem?

  • Integrate using trigonometric identities
  • Find the derivative using the definition
  • Solve by quadratic formula (general formula)
  • Simplify
  • Find the integral
  • Find the derivative
  • Factor
  • Factor by completing the square
  • Find the roots
  • Find break even points
  • Load more...
Can't find a method? Tell us so we can add it.
1

Calcular la integral

$\int\ln\left(\frac{e^{4x}-1}{e^{4x}+1}\right)dx$

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$\int\ln\left(\frac{e^{4x}-1}{e^{4x}+1}\right)dx$

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Learn how to solve problems step by step online. Solve the logarithmic equation y=ln((e^(4x)-1)/(e^(4x)+1)). Calcular la integral. The logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator. Expandir la integral \int\left(\ln\left(e^{4x}-1\right)-\ln\left(e^{4x}+1\right)\right)dx en 2 integrales usando la regla de la integral de una suma de funciones, para luego resolver cada integral por separado. La integral \int\ln\left(e^{4x}-1\right)dx da como resultado: \left(e^{4x}-1\right)\ln\left(e^{4x}-1\right)-e^{4x}+1.

Final answer to the problem

$\left(e^{4x}-1\right)\ln\left(e^{4x}-1\right)+\left(-e^{4x}-1\right)\ln\left(e^{4x}+1\right)+C_1$

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Function Plot

Plotting: $\left(e^{4x}-1\right)\ln\left(e^{4x}-1\right)+\left(-e^{4x}-1\right)\ln\left(e^{4x}+1\right)+C_1$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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