Problem 37

Question

In Problems $37-42, a, b$, and $c$ are constants and $g(x)$ is a continuous function whose derivative $g^{\prime}(x)$ is also continuous. Use substitution to evaluate each indefinite integral. $$ \int \frac{2 a x+b}{a x^{2}+b x+c} d x $$

Step-by-Step Solution

Verified

Answer

$ \ln|ax^2 + bx + c| + C $.

1Step 1: Identify the Substitution

Recognize that the structure of the integral suggests using the substitution method. Let $ u = ax^2 + bx + c $, the denominator of the integrand. This choice simplifies the derivative of $ u $ to align with the numerator.

2Step 2: Compute the Derivative of u

Find the derivative of $ u $ with respect to $ x $, which is $ \frac{du}{dx} = 2ax + b $. This aligns nicely with the numerator of the integrand.

3Step 3: Express dx in terms of du

Rearrange $ \frac{du}{dx} = 2ax + b $ to express $ dx $: \[ dx = \frac{1}{2ax + b} du \].

4Step 4: Substitute into the Integral

Substitute $ u $ and $ dx $ back into the integral: \[ \int \frac{2ax + b}{ax^2 + bx + c} dx = \int \frac{2ax + b}{u} \cdot \frac{1}{2ax + b} du \]. This simplifies to $ \int \frac{1}{u} du $.

5Step 5: Integrate with Respect to u

The integral $ \int \frac{1}{u} du $ is a standard form which results in $ \ln|u| + C $, where $ C $ is the constant of integration.

6Step 6: Substitute Back to x

Replace $ u $ with the original expression, $ ax^2 + bx + c $, giving the final answer: \[ \ln|ax^2 + bx + c| + C \].

Key Concepts

Indefinite IntegralSubstitution MethodDerivativeContinuous Function

Indefinite Integral

In calculus, an indefinite integral refers to the process of finding a function whose derivative is the given function. You might encounter it in the format of $ \int f(x) \, dx $. The result of this integral yields a family of functions, usually including a constant of integration $ C $, since differentiation erases constants.

Why do we need indefinite integrals? They essentially "reverse" the process of differentiation. If you know how a function's rate of change looks like, the indefinite integral helps you find the original function. It's all about understanding the journey from change back to the state.

The indefinite integral is symbolized by the integral sign $ \int $, followed by a function $ f(x) $, and then $ dx $, which indicates the variable of integration.
The outcome of the indefinite integral always includes a constant $ C $ because when finding the derivative, constant terms disappear.
It is a vital concept in calculus used for solving numerous mathematical problems involving area, displacement, and more.

Substitution Method

The substitution method is a technique used in integration to simplify difficult integrals by changing the variable to something more manageable. It's particularly handy when you notice that the derivative of one part of the function appears elsewhere in the integrand.

The process involves a few structured steps to make integration smoother:

First, identify a part of the integral that can be substituted with a single variable $ u $. Often, this is the portion whose derivative is also part of the integrand.
Find the derivative of $ u $ with respect to $ x $, which will help in replacing $ dx $ in the integral.
Use the substitution to transform the integral into a simpler form, integrating with respect to the new variable $ u $.

The substitution method essentially breaks a complex problem into simpler pieces, allowing for more straightforward computation.

Derivative

The derivative of a function describes its rate of change. It's a fundamental concept in calculus capturing how a function changes as its input changes. Calculating derivatives is crucial for understanding behaviors such as velocity, acceleration, and other rates of change in various scientific fields.

Consider a function $ g(x) $. Its derivative, noted as $ g^{\prime}(x) $, represents how $ g(x) $ changes for an infinitesimal change in $ x $. If $ g(x) $ is a continuous function, its derivative offers a smooth prediction of changes:

The derivative tells you the slope of the tangent line to the function at any given point.
In calculus problems, the derivative helps find the minimum and maximum values by identifying where the slope is zero (critical points).
Calculating derivatives can help approximate functions using linear equations, aiding in complex calculations.

Continuous Function

A continuous function is one where small changes in the input lead to small changes in the output. Graphically, this means you can draw the function without lifting your pencil. In calculus, these functions are significant because they allow the application of derivatives and integrals smoothly.

Continuous functions have several appealing properties in calculus:

They ensure that limits exist, which is foundational for differentiation and integration.
You can apply the Intermediate Value Theorem, which states that if a function is continuous on an interval $[a, b]$, it will take every value between $ f(a) $ and $ f(b) $.
For any continuous function $ g(x) $ with a continuous derivative $ g^{\prime}(x) $, this means the curves can be smoothly analyzed and instructions like the substitution method become applicable easily.

Understanding continuity is crucial for solving calculus problems effectively and ensures reliable results in computations.

Problem 37

Problem 38

Other exercises in this chapter

Problem 37

(a) Use integration by parts to verify the validity of the reduction formula $$ \int x^{n} e^{a x} d x=\frac{1}{a} x^{n} e^{e x}-\frac{n}{a} \int x^{n-1} e^{a x

View solution

Problem 37

Use partial fraction decompositions to evaluate each integral. $$ \int \frac{2 x-3}{(x-1)^{2}} d x $$

View solution

Problem 38

Using a spreadsheet, approximate the integral $\int_{0}^{1} \sqrt{x} d x$ using the midpoint rule and: (a) $n=5$ subintervals, (b) $n=10$ subintervals, (c

View solution

Problem 38

(a) Show that $$0 \leq \frac{1}{\sqrt{1+x^{4}}} \leq \frac{1}{x^{2}}$$ for $x>0$. (b) Use your result in (a) to show that $$\int_{1}^{\infty} \frac{1}{\sqrt{1

View solution