Question

The value of \(\int \frac{ln\,x}{(1+ln\,x)^2}dx\)=?

Answer 1

Step 1 — Spot a derivative pattern (best method)

Try to see if the integrand equals the derivative of a simple expression. Consider the function

\(F(x)=\dfrac{x}{\,1+\ln x\,}\)

Differentiate \(F(x)\) using the quotient rule (or product rule):

\[ F'(x)=\frac{(1)(1+\ln x) - x\cdot\frac{d}{dx}(\,1+\ln x\,)}{(1+\ln x)^2} =\frac{1+\ln x - x\cdot\frac{1}{x}}{(1+\ln x)^2} =\frac{1+\ln x - 1}{(1+\ln x)^2} =\frac{\ln x}{(1+\ln x)^2}. \]

We find that \(F'(x)\) exactly equals the integrand.

Step 2 — Write the antiderivative

Since the derivative of \( \dfrac{x}{1+\ln x} \) is the integrand, the indefinite integral is

\(\displaystyle \int \frac{\ln x}{(1+\ln x)^2}\,dx = \frac{x}{1+\ln x} + C\),

where \(C\) is an arbitrary constant of integration.

Step 3 — Domain & notes

The integrand involves \(\ln x\), so it is defined only for \(x>0\). Also the denominator \((1+\ln x)^2\) vanishes when \(1+\ln x=0\), i.e. at \(x=\mathrm{e}^{-1}=\tfrac{1}{e}\). Thus the integrand (and the antiderivative form) is valid on the intervals \((0,\tfrac{1}{e})\) and \((\tfrac{1}{e},\infty)\) separately. The antiderivative \(x/(1+\ln x)\) differentiates to the integrand wherever \(x>0\) and \(x\neq 1/e\).

Step 4 — Quick derivative check

Differentiate the answer to verify:

\[ \frac{d}{dx}\!\left(\frac{x}{1+\ln x}\right) =\frac{(1+\ln x)-1}{(1+\ln x)^2}=\frac{\ln x}{(1+\ln x)^2}, \]

which matches the original integrand — so the result is confirmed.

Alternate viewpoint (why this trick works)

We looked for an expression whose derivative produces a rational function in \(1+\ln x\). Noticing \(d(1+\ln x)/dx=1/x\) suggests that \(x\) multiplied by a function of \(1+\ln x\) is a natural candidate. The quotient \(x/(1+\ln x)\) is the simplest such candidate and works directly.

Final answer

\(\boxed{\displaystyle \int \frac{\ln x}{(1+\ln x)^2}\,dx = \frac{x}{1+\ln x} + C}\)