3 years ago

Does 2x=x+x just say yes or no please

1 answer:

6 0

Yes

Hope this helps!!!!

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Degger [83]

An example could be 3/4

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3 years ago

Leto [7]

Answer:its D

Step-by-step explanation:

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3 years ago

natita [175]

Answer:

see below

Step-by-step explanation:

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3 years ago

Andreyy89

Answer:

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}$ .

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}$ .

In other words,

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}$ .

Apply $u$ -substitution on the integral $\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}$ :

Let $u = 1 + e^{x}$ . $u > 1$ .

$du = e^{x}\cdot dx$ .

$\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C$ .

Therefore

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}$ .

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3 years ago

Lena [83]

The answer is radical 32 or 5.656854249

5 0

4 years ago