Question Evaluate the integral $\int_{e^{3 x}}^{e^{4 x}} \frac{1}{x \ln x+x} dx$ .

Studdy Solution

STEP 1

Assumptions
1. The integral to be evaluated is $\int_{e^{3x}}^{e^{4x}} \frac{1}{x\ln(x) + x} dx$ .
2. The base of the logarithm $\log_e$ is the natural logarithm, which can be written as $\ln$ .
3. The variable of integration is $x$ , and the limits of integration are functions of $x$ .
4. We will use substitution to simplify the integral.

STEP 2

Identify a substitution that will simplify the integral. Since the denominator has the term $x\ln(x) + x$ , we can notice that the derivative of $\ln(x)$ is $\frac{1}{x}$ , which is present in the numerator. This suggests that we can use substitution with $u = \ln(x)$ , which implies $x = e^u$ .

STEP 3

Differentiate $u = \ln(x)$ with respect to $x$ to find $du$ .
$du = \frac{1}{x}dx$

STEP 4

Rearrange the expression for $du$ to solve for $dx$ .
$dx = x du$

STEP 5

Substitute $x = e^u$ into the expression for $dx$ to express $dx$ entirely in terms of $u$ .
$dx = e^u du$

STEP 6

Now we need to change the limits of integration. The original lower limit is $e^{3x}$ , so when $x = e^{3x}$ , $u = \ln(e^{3x}) = 3x$ . Similarly, for the upper limit, when $x = e^{4x}$ , $u = \ln(e^{4x}) = 4x$ .

STEP 7

Notice that the limits of integration are still in terms of $x$ . To express them in terms of $u$ , we need to recognize that $u$ is a function of $x$ and not a multiple of $x$ . Therefore, we need to find $u$ when $x$ is $e^{3x}$ and $e^{4x}$ , which is not straightforward. This suggests that a direct substitution may not be the best approach.

STEP 8

Let's reconsider our substitution strategy. We can try a different substitution that might simplify the integral. Let's let $u = x\ln(x) + x$ , which is the entire denominator of the integrand.

STEP 9

Differentiate $u = x\ln(x) + x$ with respect to $x$ to find $du$ .
$du = (1\cdot\ln(x) + x\cdot\frac{1}{x} + 1)dx$

STEP 10

Simplify the expression for $du$ .
$du = (\ln(x) + 1 + 1)dx$ $du = (\ln(x) + 2)dx$

STEP 11

Rearrange the expression for $du$ to solve for $dx$ .
$dx = \frac{du}{\ln(x) + 2}$

STEP 12

Notice that we have $\ln(x)$ in the expression for $dx$ , which is not in terms of $u$ . We need to express $\ln(x)$ in terms of $u$ to perform the substitution. From our substitution $u = x\ln(x) + x$ , we can solve for $\ln(x)$ in terms of $u$ and $x$ .

STEP 13

Rearrange the substitution equation to solve for $\ln(x)$ .
$\ln(x) = \frac{u - x}{x}$

STEP 14

Now, we need to express $x$ in terms of $u$ to eliminate $x$ from the equation. However, the equation $u = x\ln(x) + x$ does not allow for an easy expression of $x$ in terms of $u$ . This suggests that our substitution choice may not be the best approach.

STEP 15

Let's return to our original substitution idea, $u = \ln(x)$ , and carefully re-evaluate the limits of integration. We need to express the limits of integration in terms of $u$ correctly.

STEP 16

For the lower limit of integration, when $x = e^{3x}$ , we need to find the value of $u$ that corresponds to this $x$ . Since $u = \ln(x)$ , we have $u = \ln(e^{3x})$ . However, this is still in terms of $x$ , so we need to find a way to express $u$ in terms of itself, not in terms of $x$ .

STEP 17

Recognize that the limits of integration are not constants but functions of $x$ . This means we cannot directly substitute the limits as we would with constant limits. We need to find a different approach to handle the variable limits.

STEP 18

Since the limits of integration are functions of $x$ , we can use the Leibniz rule for integration, which allows us to differentiate under the integral sign. Let's define a new function $F(x)$ that represents our integral with variable limits.
$F(x) = \int_{e^{3x}}^{e^{4x}} \frac{1}{x\ln(x) + x} dx$

STEP 19

Differentiate $F(x)$ with respect to $x$ using the Leibniz rule.
$F'(x) = \frac{d}{dx} \left( \int_{e^{3x}}^{e^{4x}} \frac{1}{x\ln(x) + x} dx \right)$

STEP 20

Apply the Leibniz rule, which states that the derivative of the integral with respect to $x$ is the difference of the integrand evaluated at the upper limit times the derivative of the upper limit minus the integrand evaluated at the lower limit times the derivative of the lower limit.
$F'(x) = \frac{1}{e^{4x}\ln(e^{4x}) + e^{4x}} \cdot \frac{d}{dx}(e^{4x}) - \frac{1}{e^{3x}\ln(e^{3x}) + e^{3x}} \cdot \frac{d}{dx}(e^{3x})$

STEP 21

Differentiate the upper and lower limits with respect to $x$ .
$\frac{d}{dx}(e^{4x}) = 4e^{4x}$ $\frac{d}{dx}(e^{3x}) = 3e^{3x}$

STEP 22

Substitute the derivatives of the limits into the expression for $F'(x)$ .
$F'(x) = \frac{1}{e^{4x}\ln(e^{4x}) + e^{4x}} \cdot 4e^{4x} - \frac{1}{e^{3x}\ln(e^{3x}) + e^{3x}} \cdot 3e^{3x}$

STEP 23

Simplify the expression by canceling out terms.
$F'(x) = \frac{4}{4x + 1} - \frac{3}{3x + 1}$

STEP 24

Now, we need to integrate $F'(x)$ with respect to $x$ to find $F(x)$ .
$F(x) = \int \left( \frac{4}{4x + 1} - \frac{3}{3x + 1} \right) dx$

STEP 25

Split the integral into two separate integrals.
$F(x) = \int \frac{4}{4x + 1} dx - \int \frac{3}{3x + 1} dx$

STEP 26

Use a simple substitution for each integral. For the first integral, let $v = 4x + 1$ , and for the second integral, let $w = 3x + 1$ .

STEP 27

Differentiate $v$ and $w$ with respect to $x$ .
$dv = 4dx$ $dw = 3dx$

STEP 28

Solve for $dx$ in terms of $dv$ and $dw$ .
$dx = \frac{dv}{4}$ $dx = \frac{dw}{3}$

STEP 29

Substitute $dv$ and $dw$ into the integrals and integrate.
$F(x) = \int \frac{1}{v} \frac{dv}{4} - \int \frac{1}{w} \frac{dw}{3}$
$F(x) = \frac{1}{4} \int \frac{1}{v} dv - \frac{1}{3} \int \frac{1}{w} dw$
$F(x) = \frac{1}{4} \ln|v| - \frac{1}{3} \ln|w| + C$

STEP 30

Substitute back for $v$ and $w$ to get $F(x)$ in terms of $x$ .
$F(x) = \frac{1}{4} \ln|4x + 1| - \frac{1}{3} \ln|3x + 1| + C$
This function $F(x)$ represents the antiderivative of the original integral. However, we must acknowledge that the original problem posed an improper integral with variable limits of integration, which cannot be evaluated using standard methods. The steps provided here lead to an antiderivative of a related function, but they do not solve the original problem as stated. The original integral requires a more sophisticated approach or clarification of the problem statement.

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