Integral Exercise 8

We are asked to evaluate the following integral:

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

To simplify the integrand, we’ll use the method of partial fraction decomposition.

The denominator is the product of two distinct linear factors, each with multiplicity one, so we can express the integrand as:

$$ \frac{1}{(1+x)x} = \frac{A}{1+x} + \frac{B}{x} $$

where \( A \) and \( B \) are constants to be determined.

Bringing the right-hand side to a common denominator:

$$ \frac{A}{1+x} + \frac{B}{x} = \frac{Ax + B(1+x)}{(1+x)x} $$

Expanding the numerator:

$$ = \frac{Ax + B + Bx}{(1+x)x} = \frac{(A + B)x + B}{(1+x)x} $$

Since the numerators must be equal, we compare coefficients:

$$ \begin{cases} B = 1 \\ A + B = 0 \end{cases} $$

Explanation. On the left-hand side, there is no term in \( x \), so its coefficient is 0; on the right-hand side, the coefficient of \( x \) is \( A + B \), which gives us the equation \( A + B = 0 \). The constant term on the left is 1, and on the right it is \( B \), so \( B = 1 \).

Solving the system yields:

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

Substituting these values back into the decomposition, we obtain:

$$ \frac{1}{(1+x)x} = \frac{-1}{1+x} + \frac{1}{x} $$

Thus, the original integral becomes:

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

By rewriting the product in the denominator as a sum of simple fractions, we've reduced the problem to the evaluation of two elementary integrals:

$$ - \int \frac{1}{1+x} \ dx + \int \frac{1}{x} \ dx $$

Each integral is standard:

The first is the natural logarithm of \( 1 + x \):

$$ -\log(1+x) + \int \frac{1}{x} \ dx + c $$

The second is the natural logarithm of \( x \):

$$ -\log(1+x) + \log(x) + c $$

Therefore, the final result is:

$$ \log(x) - \log(1+x) + c $$

And that completes the solution.