elnx 8) fex dx g) √1-e²t dt (Hint: let u = e¹) h) fx dx dx (let u = x¹/2)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please solve for the integrals using substitution
### Integral Problems

#### Problem g
\[ \int \frac{e^{\ln x}}{x} \, dx \]

#### Problem h
\[ \int \frac{\ln x}{\sqrt{x}} \, dx \quad \text{(let } u = x^{1/2} \text{)} \]

#### Problem i
\[ \int \frac{\frac{e^t}{\sqrt{1 - e^{2t}}}}{1-e^{2t}} \, dt \quad \text{(Hint: let } u = e^t\text{)} \]

### Detailed Explanations
- **Problem g**: The integral involves the natural log function and the exponential function. Simplifying using properties of logarithms and exponentials could be a useful first step.

- **Problem h**: Substitution is suggested where \( u = x^{1/2} \). This transforms the integral into a more manageable form, focusing on powers of \( x \) and their derivatives.

- **Problem i**: The hint suggests using substitution again, where \( u = e^t \). This is likely to simplify the expression under the square root and the fraction, helping solve the integral more easily.
Transcribed Image Text:### Integral Problems #### Problem g \[ \int \frac{e^{\ln x}}{x} \, dx \] #### Problem h \[ \int \frac{\ln x}{\sqrt{x}} \, dx \quad \text{(let } u = x^{1/2} \text{)} \] #### Problem i \[ \int \frac{\frac{e^t}{\sqrt{1 - e^{2t}}}}{1-e^{2t}} \, dt \quad \text{(Hint: let } u = e^t\text{)} \] ### Detailed Explanations - **Problem g**: The integral involves the natural log function and the exponential function. Simplifying using properties of logarithms and exponentials could be a useful first step. - **Problem h**: Substitution is suggested where \( u = x^{1/2} \). This transforms the integral into a more manageable form, focusing on powers of \( x \) and their derivatives. - **Problem i**: The hint suggests using substitution again, where \( u = e^t \). This is likely to simplify the expression under the square root and the fraction, helping solve the integral more easily.
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