/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 79

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is an integer. $$\int \frac{x}{a x+b} d x \text { (Use } u=a x+b$$

Short Answer

Expert verified
Question: Determine the integral of the function \(f(x) = \frac{x}{ax + b}\) using the substitution \(u = ax + b\). Answer: The integral of the function \(f(x) = \frac{x}{ax + b}\) using the substitution \(u = ax + b\) is \(\int \frac{x}{a x+b} d x = \frac{1}{a^2} \left[(ax+b) - b \ln|ax+b|\right] + C\).

Step by step solution

01

Substitute u into the function

Using the substitution method, let's substitute \(u = ax + b\). Now, we need to find the differential of \(u\), which is \(du\). Differentiate both sides of the equation \(u = ax + b\) with respect to \(x\): $$\frac{du}{dx} = a$$ Now, solve for \(dx\): $$dx = \frac{du}{a}$$ Next, we replace \(dx\) and \(ax+b\) in the original integral: $$\int \frac{x}{a x+b} d x = \int \frac{x}{u} \cdot \frac{du}{a}$$
02

Simplify the integral

Now, we simplify the integral by factoring out the term \(x\) and the constant \(\frac{1}{a}\): $$\int \frac{x}{u} \cdot \frac{du}{a} = \frac{1}{a} \int x \cdot \frac{1}{u} du$$
03

Express x in terms of u

We need to express \(x\) in terms of \(u\) from the substitution relation \(u = ax + b\). Then, we will put it back into the integral: $$x = \frac{u - b}{a}$$ So our integral becomes: $$\frac{1}{a} \int \frac{u - b}{a} \cdot \frac{1}{u} du$$
04

Simplify the integral again and integrate

Now, we simplify the integral further and integrate with respect to \(u\): $$\frac{1}{a} \int \left( \frac{u}{u} - \frac{b}{u} \right) \frac{1}{a} du = \frac{1}{a^2} \int \left(1 - \frac{b}{u}\right) du$$ Now, we can integrate each term individually: $$\frac{1}{a^2} \left[\int 1 du - \int \frac{b}{u} du\right] = \frac{1}{a^2} \left[u - b \ln|u|\right] + C$$
05

Replace u with original terms

Now that we have integrated with respect to \(u\), we can substitute back the original terms for \(u\) to find the solution in terms of \(x\): $$\frac{1}{a^2} \left[u - b \ln|u|\right] + C = \frac{1}{a^2} \left[(ax+b) - b \ln|ax+b|\right] + C$$ So the final answer is: $$\int \frac{x}{a x+b} d x = \frac{1}{a^2} \left[(ax+b) - b \ln|ax+b|\right] + C$$

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