Integral Calculation Exercise 40

We are asked to evaluate the following integral:

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

The integrand is a rational function, and since the degree of the numerator is less than that of the denominator, we don’t need to perform polynomial division.

Let’s start by factoring the denominator to express it as a product of two linear terms:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx $$

To proceed, we apply the method of partial fraction decomposition.

Since both roots are distinct and of multiplicity one, we can decompose the expression as follows, where A and B are constants to be determined:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = \int \frac{A}{x} + \frac{B}{x+3} \ dx $$

Bringing the right-hand side to a common denominator:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = \int \frac{A(x+3) + Bx}{x \cdot (x+3)} \ dx $$

Expanding the numerator gives:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = \int \frac{(A + B)x + 3A}{x \cdot (x+3)} \ dx $$

Since both sides have the same denominator, we can equate the numerators:

x - 1 = (A + B)x + 3A

This leads to a system of equations:

$$ \begin{cases} A + B = 1 \\ 3A = -1 \end{cases} $$

Solving the second equation gives \( A = -\frac{1}{3} \), which we substitute into the first:

$$ -\frac{1}{3} + B = 1 \Rightarrow B = \frac{4}{3} $$

So we’ve found the constants: A = -1/3 and B = 4/3.

Substituting these back into the integral:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = \int \left( \frac{-1}{3x} + \frac{4}{3(x+3)} \right) dx $$

Now, using the linearity of the integral, we split it into two parts:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = -\frac{1}{3} \int \frac{1}{x} \ dx + \frac{4}{3} \int \frac{1}{x+3} \ dx $$

Both integrals are standard:

$$ \int \frac{1}{x} \ dx = \log|x| + C \quad \text{and} \quad \int \frac{1}{x+3} \ dx = \log|x+3| + C $$

Putting it all together:

$$ \int \frac{x-1}{x \cdot (x+3)} \ dx = -\frac{1}{3} \log|x| + \frac{4}{3} \log|x+3| + C $$

And that completes the evaluation of the integral.

And so on.