Problem 27

Question

$$ \text { In Problems } 27-36, \text { evaluate each integral. } $$ $$ \int \frac{1}{(x-3)(x+2)} d x $$

Step-by-Step Solution

Verified

Answer

$ \frac{3}{5} \ln\left| \frac{x-3}{x+2} \right| + C $

1Step 1: Setup Partial Fraction Decomposition

Given the integral $ \int \frac{1}{(x-3)(x+2)} \, dx $, we first express the integrand using partial fraction decomposition. We know that $ \frac{1}{(x-3)(x+2)} $ can be written as $ \frac{A}{x-3} + \frac{B}{x+2} $. Our goal is to find the constants $ A $ and $ B $.

2Step 2: Solve for Partial Fraction Coefficients

To find $ A $ and $ B $, we equate: $ 1 = A(x+2) + B(x-3) $. Expanding and combining like terms, we get: $ A x + 2A + B x - 3B = 1 $. This simplifies to: $ (A + B)x + (2A - 3B) = 1 $. We equate the coefficients with those on the right: $ A + B = 0 $ and $ 2A - 3B = 1 $. Solve these equations to find $ A = \frac{3}{5} $ and $ B = -\frac{3}{5} $.

3Step 3: Integrate Partial Fractions

Now we integrate the partial fractions: $ \int \left( \frac{\frac{3}{5}}{x-3} - \frac{\frac{3}{5}}{x+2} \right) dx $. Separate the integrals to get: $ \frac{3}{5} \int \frac{1}{x-3} \, dx - \frac{3}{5} \int \frac{1}{x+2} \, dx $.

4Step 4: Find Antiderivatives

The antiderivative of $ \int \frac{1}{x-3} \, dx $ is $ \ln|x-3| $, and the antiderivative of $ \int \frac{1}{x+2} \, dx $ is $ \ln|x+2| $. So, we have: $ \frac{3}{5} \ln|x-3| - \frac{3}{5} \ln|x+2| + C $.

5Step 5: Simplify the Expression

Combine the logarithms: $ \frac{3}{5} (\ln|x-3| - \ln|x+2|) + C $. Using the properties of logarithms, this simplifies to: $ \frac{3}{5} \ln\left| \frac{x-3}{x+2} \right| + C $. Thus, the evaluated integral is: $ \frac{3}{5} \ln\left| \frac{x-3}{x+2} \right| + C $.

Key Concepts

Partial Fraction DecompositionAntiderivativeLogarithmic Integration

Partial Fraction Decomposition

When dealing with calculus integrals, it's common to encounter rational expressions that can be simplified for easier integration. One powerful technique is partial fraction decomposition. This transforms complex rational expressions into a sum of simpler fractions.
By breaking down the original fraction into simpler parts, integrating becomes straightforward.Here’s how it works:

Look at the integral: $ \int \frac{1}{(x-3)(x+2)} \, dx $. This has a denominator that can be factored into linear parts $(x-3)$ and $(x+2)$.
Express $ \frac{1}{(x-3)(x+2)} $ as $ \frac{A}{x-3} + \frac{B}{x+2} $, where $ A $ and $ B $ are constants to be determined.

The ultimate goal is to find these constants so that the new expression matches the original one. Equate coefficients and solve the resulting system of equations. This decomposition simplifies the process of taking the antiderivative later.

Antiderivative

Integrating a function means finding its antiderivative. This reverses the process of differentiation. In our exercise, after setting up the partial fractions, we need to take the antiderivative of each part separately. For our integral:
$ \int \frac{1}{(x-3)(x+2)} \, dx $ becomes two separate integrals:

$ \frac{3}{5} \int \frac{1}{x-3} \, dx $
- $ \frac{3}{5} \int \frac{1}{x+2} \, dx $

These are easier to handle since they match a known antiderivative form. To find antiderivatives:

The antiderivative of $ \frac{1}{x-3} $ is $ \ln|x-3| $
The antiderivative of $ \frac{1}{x+2} $ is $ \ln|x+2| $

So, the antiderivatives are natural logarithms of the expressions obtained from partial fraction decomposition.

Logarithmic Integration

Logarithmic integration appears often in calculus, especially with denominators of linear terms. This type of integration leverages the properties of logarithms to simplify and solve expressions.In our specific exercise: