Given y = (1+ ln x)/(x-x ln x), verify the equality 2x²

Given $y = \dfrac{1+\ln x}{x-x\ln x}$, verify the equality $2x^{2}\,dy = \bigl(x^{2}y^{2}+1\bigr)\,dx$.

Step 1: Factor the denominator

\[ y = \frac{1+\ln x}{x-x\ln x} = \frac{1+\ln x}{x\bigl(1-\ln x\bigr)}. \]

The domain is $x>0,\ x\neq e$ (the denominator vanishes at $\ln x = 1$).

Step 2: Differentiate by the quotient rule

Take $u = 1+\ln x$ and $v = x(1-\ln x)$. Then $u' = \dfrac1x$, and for $v$ the product rule gives

\[ v' = \bigl(1-\ln x\bigr) + x\cdot\left(-\frac1x\right) = 1-\ln x-1 = -\ln x. \]

\[ y' = \frac{u'v-uv'}{v^{2}} = \frac{\dfrac1x\cdot x\bigl(1-\ln x\bigr)-\bigl(1+\ln x\bigr)\bigl(-\ln x\bigr)}{x^{2}\bigl(1-\ln x\bigr)^{2}}. \]

Step 3: Simplify the numerator

\[ \bigl(1-\ln x\bigr)+\ln x\bigl(1+\ln x\bigr) = 1-\ln x+\ln x+\ln^{2}x = 1+\ln^{2}x. \]

\[ y' = \frac{1+\ln^{2}x}{x^{2}\bigl(1-\ln x\bigr)^{2}}. \tag{1} \]

The sign is positive: the two $\ln x$ terms cancel rather than reinforce, and the result is a sum of squares over a square, hence $y'>0$ on the whole domain.

Step 4: Compute $x^{2}y^{2}+1$

\[ y^{2} = \frac{\bigl(1+\ln x\bigr)^{2}}{x^{2}\bigl(1-\ln x\bigr)^{2}} \;\Longrightarrow\; x^{2}y^{2} = \frac{\bigl(1+\ln x\bigr)^{2}}{\bigl(1-\ln x\bigr)^{2}}. \]

Adding $1$ and using $(1+L)^{2}+(1-L)^{2} = 2+2L^{2}$ with $L = \ln x$: \[x^{2}y^{2}+1 = \frac{\bigl(1+\ln x\bigr)^{2}+\bigl(1-\ln x\bigr)^{2}}{\bigl(1-\ln x\bigr)^{2}} = \frac{2\bigl(1+\ln^{2}x\bigr)}{\bigl(1-\ln x\bigr)^{2}}. \tag{2}\]

Step 5: Compare the two sides

Left. Using (1), the factor $x^{2}$ cancels completely:

\[ 2x^{2}\,dy = 2x^{2}y'\,dx = 2x^{2}\cdot\frac{1+\ln^{2}x}{x^{2}\bigl(1-\ln x\bigr)^{2}}\,dx = \frac{2\bigl(1+\ln^{2}x\bigr)}{\bigl(1-\ln x\bigr)^{2}}\,dx. \]

Right. By (2),

\[ \bigl(x^{2}y^{2}+1\bigr)\,dx = \frac{2\bigl(1+\ln^{2}x\bigr)}{\bigl(1-\ln x\bigr)^{2}}\,dx. \]

The two expressions are identical, which proves the claim.

Step 6: Verification

\[x = 1:\quad y = 1,\ y' = 1; \qquad 2\cdot1\cdot1 = 2 \quad\text{and}\quad 1\cdot1+1 = 2 \;\;\checkmark\]

\[x = e^{2}:\quad y = \frac{3}{e^{2}(-1)} \approx -0.406006,\ y' = \frac{5}{e^{4}} \approx 0.091578\]

\[2e^{4}\cdot\frac{5}{e^{4}} = 10 \quad\text{and}\quad e^{4}\cdot\frac{9}{e^{4}}+1 = 10 \;\;\checkmark\]