How do you integrate #int ln(e^(2x-1))dx#?

Question

Charles Anctil · Accepted Answer

# x^2 - x +c #

Apply your knowledge of logarithms.

# log_a a^f(x) -= f(x) ln_a a -= f(x) #

Hence #ln e^f(x) -= f(x) #

#=> int ln (e^(2x-1) ) dx = int (2x-1) dx#

Apply the rule of reverse power:

# = x^2 -x + c #

Ezekiel Alexander · Answer

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$$

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$$

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$$
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$$

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$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

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$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

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$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(To integrate \( \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrateTo integrate $\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2To integrate \( \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2x-To integrate \( \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $ To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2x-1)$To integrate $ \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $ 2To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2x-1)$ withTo integrate $ \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $ 2x -To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2x-1)$ with respect toTo integrate $ \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $ 2x - To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

Now, we can integrate $(2x-1)$ with respect to $xTo integrate \( \int \ln(e^{2x-1}) \, dx $, we can use the properties of logarithms and the integral of exponential functions. First, we simplify the expression inside the logarithm using the property $ \ln(e^a) = a $. This gives us $ \ln(e^{2x-1}) = 2x - 1 $. Now, we integrate $ 2x - 1To integrate \(\int \ln(e^{2x-1}) \, dx$, we can use the property of logarithms that $\ln(e^u) = u$. So, we have:

$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$

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$$
\int \ln(e^{2x-1}) \, dx = \int (2x-1) \, dx
$$