Let’s solve these integrals step by step.

1. $\int \frac{1}{1-x} \, dx$

This is a standard integral of the form:

$\int \frac{1}{1-x} \, dx = -\ln|1-x| + C$

Solution: $-\ln|1-x| + C$

2. $\int \frac{5}{2x+7} \, dx$

Substitute $u = 2x+7$, then $du = 2 \, dx$, or $dx = \frac{du}{2}$.

Rewriting: $\int \frac{5}{2x+7} \, dx = \int \frac{5}{u} \cdot \frac{1}{2} \, du = \frac{5}{2} \int \frac{1}{u} \, du = \frac{5}{2} \ln|u| + C$

Back-substitute $u = 2x+7$: $\frac{5}{2} \ln|2x+7| + C$

3. $\int_{-1}^2 \frac{2x}{x^2+4} \, dx$

Substitute $u = x^2+4$, then $du = 2x \, dx$.

Rewriting: $\int \frac{2x}{x^2+4} \, dx = \int \frac{1}{u} \, du = \ln|u| + C$

Back-substitute $u = x^2+4$: $\ln(x^2+4) \bigg|_{-1}^2$

Evaluate the limits: $\ln(2^2 + 4) - \ln((-1)^2 + 4) = \ln(8) - \ln(5) = \ln\left(\frac{8}{5}\right)$

Solution: $\ln\left(\frac{8}{5}\right)$

4. $\int_0^{\ln 2} e^{-3x} \, dx$

Integrating $e^{-3x}$: $\int e^{-3x} \, dx = \frac{e^{-3x}}{-3} + C = -\frac{1}{3} e^{-3x} + C$

Evaluate the definite integral: $-\frac{1}{3} e^{-3x} \bigg|_0^{\ln 2} = -\frac{1}{3} \left(e^{-3 \ln 2} - e^{0}\right)$

Simplify: $e^{-3 \ln 2} = (e^{\ln 2})^{-3} = 2^{-3} = \frac{1}{8}$ $-\frac{1}{3} \left(\frac{1}{8} - 1\right) = -\frac{1}{3} \left(-\frac{7}{8}\right) = \frac{7}{24}$

Solution: $\frac{7}{24}$

5. $\int \frac{e^{1/x}}{x^2} \, dx$

Substitute $u = \frac{1}{x}$, then $du = -\frac{1}{x^2} dx$.

Rewriting: $\int \frac{e^{1/x}}{x^2} \, dx = -\int e^u \, du = -e^u + C$

Back-substitute $u = \frac{1}{x}$: $-e^{1/x} + C$

6. $\int (e^x+1)^3 \cdot e^x \, dx$

Substitute $u = e^x+1$, then $du = e^x \, dx$.

Rewriting: $\int (e^x+1)^3 \cdot e^x \, dx = \int u^3 \, du = \frac{u^4}{4} + C$

Back-substitute $u = e^x+1$: $\frac{(e^x+1)^4}{4} + C$

Do you want detailed steps for any of these? Here are 5 related questions:

1. How do you handle definite integrals with logarithmic limits?
2. Can substitution always simplify complex integrals?
3. What are common integration tricks for exponential functions?
4. How do you determine the bounds after substitution?
5. Can improper integrals converge if they involve $\ln|x|$?

Tip: Always check substitution bounds carefully when solving definite integrals to avoid missing changes in limits.