Integral Calculation Exercise 6

We are asked to evaluate the following definite integral:

$$ \int_1^{\sqrt{e}} \frac{\log(x) - 2}{x[\log^2(x) - 1]} \ dx $$

We'll solve this by making an appropriate substitution. Let’s introduce the auxiliary variable:

$$ t = \log(x) $$

Taking differentials on both sides gives:

$$ dt = \frac{1}{x} \ dx $$

We now rewrite the integral in terms of the new variable $ t $, noting that $ t = \log(x) $.

The limits of integration transform as follows:

$$ \log(1) = 0, \quad \log(\sqrt{e}) = \frac{1}{2} $$

So the integral becomes:

$$ \int_0^{\frac{1}{2}} \frac{\log(x) - 2}{x[\log^2(x) - 1]} \ dx $$

Substituting $ t = \log(x) $ into the integrand:

$$ \int_0^{\frac{1}{2}} \frac{t - 2}{x(t^2 - 1)} \ dx $$

Now using the differential $ dt = \frac{1}{x} dx $, we obtain:

$$ \int_0^{\frac{1}{2}} \frac{t - 2}{t^2 - 1} \ dt $$

Next, we factor the denominator using the identity $ t^2 - 1 = (t - 1)(t + 1) $:

$$ \int_0^{\frac{1}{2}} \frac{t - 2}{(t - 1)(t + 1)} \ dt $$

We now apply the method of partial fractions to decompose the integrand:

$$ \frac{t - 2}{(t - 1)(t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1} $$

Multiplying both sides by the denominator and simplifying:

$$ t - 2 = A(t + 1) + B(t - 1) = t(A + B) + (A - B) $$

Matching coefficients, we get the system:

$$ A + B = 1 \\ A - B = -2 $$

Solving this system yields:

$$ A = -\frac{1}{2}, \quad B = \frac{3}{2} $$

Substituting back, the integral becomes:

$$ \int_0^{\frac{1}{2}} \left( \frac{-\frac{1}{2}}{t - 1} + \frac{\frac{3}{2}}{t + 1} \right) \ dt $$

We now apply the linearity of the integral:

$$ -\frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{t - 1} \ dt + \frac{3}{2} \int_0^{\frac{1}{2}} \frac{1}{t + 1} \ dt $$

Both are elementary integrals. Their antiderivatives are $ \log|t - 1| $ and $ \log|t + 1| $, respectively:

$$ -\frac{1}{2} \left[ \log|t - 1| \right]_0^{\frac{1}{2}} + \frac{3}{2} \left[ \log|t + 1| \right]_0^{\frac{1}{2}} $$

Note. We omit the constant of integration since this is a definite integral.

Now evaluate the two logarithmic expressions:

$$ -\frac{1}{2} \left( \log\left| \frac{1}{2} - 1 \right| - \log| -1 | \right) + \frac{3}{2} \left( \log\left| \frac{3}{2} \right| - \log|1| \right) $$

Since $ \log|1| = 0 $ and $ \log|-1| = \log(1) = 0 $, we simplify to:

$$ -\frac{1}{2} \log\left( \frac{1}{2} \right) + \frac{3}{2} \log\left( \frac{3}{2} \right) $$

Therefore, the value of the definite integral is:

$$ \frac{3}{2} \cdot \log\left( \frac{3}{2} \right) - \frac{1}{2} \cdot \log\left( \frac{1}{2} \right) $$

And that concludes the computation.