The integral in the image is:

$\int_1^{e^4} \frac{\ln x}{x^7} \, dx$

To solve this, let's proceed as follows:

Step 1: Substitution

Let: $u = \ln x \quad \text{so that} \quad du = \frac{1}{x} \, dx$

This substitution also changes the limits:

• When $x = 1$, $u = \ln 1 = 0$.
• When $x = e^4$, $u = \ln(e^4) = 4$.

Rewriting the integral in terms of $u$: $\int_1^{e^4} \frac{\ln x}{x^7} \, dx = \int_0^4 \frac{u}{e^{6u}} \, du$

Step 2: Simplify the integrand

Since $e^{6u} = (e^u)^6$, the integrand becomes: $\int_0^4 \frac{u}{e^{6u}} \, du = \int_0^4 u e^{-6u} \, du$

Step 3: Integration by parts

Using integration by parts, let:

• $v = u \quad \text{and} \quad dw = e^{-6u} \, du$
• Then $dv = du \quad \text{and} \quad w = \int e^{-6u} \, du = -\frac{1}{6} e^{-6u}$.

The formula for integration by parts is: $\int v \, dw = vw - \int w \, dv$

Applying this: $\int u e^{-6u} \, du = \left[u \left(-\frac{1}{6} e^{-6u}\right)\right]_0^4 - \int \left(-\frac{1}{6} e^{-6u}\right) \, du$

Step 4: Evaluate each term

1. First term: $\left[u \left(-\frac{1}{6} e^{-6u}\right)\right]_0^4 = \left(-\frac{1}{6} \cdot 4 e^{-24}\right) - \left(-\frac{1}{6} \cdot 0 e^0\right) = -\frac{4}{6} e^{-24} + 0 = -\frac{2}{3} e^{-24}.$

2. Second term: $\int \left(-\frac{1}{6} e^{-6u}\right) \, du = -\frac{1}{6} \int e^{-6u} \, du = -\frac{1}{6} \left(-\frac{1}{6} e^{-6u}\right) = \frac{1}{36} e^{-6u}.$ Evaluating from 0 to 4: $\left[\frac{1}{36} e^{-6u}\right]_0^4 = \frac{1}{36} e^{-24} - \frac{1}{36} e^0 = \frac{1}{36} e^{-24} - \frac{1}{36}.$

Step 5: Combine results

Adding the two parts together: $\int_1^{e^4} \frac{\ln x}{x^7} \, dx = -\frac{2}{3} e^{-24} + \frac{1}{36} e^{-24} - \frac{1}{36}.$

Simplify the coefficients of $e^{-24}$: $-\frac{2}{3} + \frac{1}{36} = -\frac{24}{36} + \frac{1}{36} = -\frac{23}{36}.$

So the final result is: $\int_1^{e^4} \frac{\ln x}{x^7} \, dx = -\frac{23}{36} e^{-24} - \frac{1}{36}.$

Let me know if you'd like further clarification or a step-by-step walkthrough!

Related Questions:

1. How is the substitution $u = \ln x$ derived in integrals involving logarithmic terms?
2. What are the key conditions for applying integration by parts effectively?
3. How does the exponential decay of $e^{-6u}$ affect the integral's convergence?
4. Can this result be approximated numerically for practical purposes?
5. Are there alternative techniques to solve this integral?

Tip: Always verify limits of substitution when dealing with definite integrals to ensure accuracy.