Double Integral of 1/x Between x = y and x = y^2

The inner integral collapses to a single logarithm

On $1 \le y \le 2$ the curve $x = y^2$ lies to the right of $x = y$, and the whole region sits in $x \ge 1$. So $\frac{1}{x}$ is continuous there and nothing is singular.

That collapse is the point of the problem. What survives is the standard integral of $\ln y$, done by parts with $u = \ln y$ and $dv = dy$.

Numerically $2\ln 2 - 1 \approx 0.386294$.

The mistake: a difference of logs is not a log of a difference

$\ln\!\left(y^2\right) - \ln y$ is $\ln\!\left(\frac{y^2}{y}\right) = \ln y$. It is not $\ln\!\left(y^2 - y\right)$. The subtraction happens to the logarithms, not to their arguments.

The faulty version is not merely a different expression, it is undefined at $y = 1$, where $y^2 - y = 0$. A formula that blows up at an endpoint the region genuinely includes is a signal to back up one step.

• $\ln a - \ln b = \ln\!\left(\frac{a}{b}\right)$, never $\ln(a-b)$.
• $\ln\!\left(y^2\right) = 2\ln y$ for $y > 0$, which is where the factor of $2$ in the answer comes from.
• $\int \frac{dx}{x} = \ln|x|$, and the absolute value is safe to drop here only because $x \ge 1$ throughout the region.

The other slip is losing the $-y$ in $\int \ln y\,dy = y\ln y - y$. Without it the answer would read $2\ln 2 \approx 1.386$, more than three times too large.

Where the region is, and why 1/x is safe here

At $y = 1$ the two curves meet at the single point $x = 1$, so the region tapers to a corner. It widens as $y$ increases, reaching $2 \le x \le 4$ at $y = 2$.

Because $x$ never approaches $0$, the integrand stays between $\frac14$ and $1$ on the region. Multiplying those bounds by the area brackets the answer between $\frac14 \cdot \frac56 \approx 0.208$ and $1 \cdot \frac56 \approx 0.833$, and $0.386294$ sits inside.

Had the region reached $x = 0$, the same integrand would have made the integral improper, and none of this bracketing would apply.