Question

the value of ∫\(\frac{dx}{x(1+log\,x)^3}\) is _______

• A. \(-\frac{x}{(1+log\,x)^2}\)\(+c\)
• B. \(-\frac{x}{(1+log\,x)^2}+c\)
• C. \(-\frac{1}{2(1+log\,x)^2}+c\)
• D. \(\frac{2}{(1+log\,x)^2}+c\)

Answer 1

The Correct Option is C

To solve the integral \(\int \frac{dx}{x(1+\log\,x)^3}\), we can use a substitution method. Let \(t = 1 + \log x\), then \(dt = \frac{1}{x} dx\) or \(dx = x\,dt = e^{t-1}\,dt\). Therefore, the integral becomes:

\[\int \frac{e^{t-1}\,dt}{e^{t}(t)^3} = \int \frac{dt}{e\,t^3}\].

This simplifies to \(\frac{1}{e} \int t^{-3}\,dt\).

Now, integrate \(t^{-3}\,dt\):

\[\int t^{-3}\,dt = \frac{t^{-2}}{-2} = -\frac{1}{2t^2} + C\], where \(C\) is the constant of integration.

Thus, the original integral evaluates to:

\[-\frac{1}{2e}\frac{1}{t^2} + C\].

Substitute back \(t = 1 + \log x\) to get:

\[-\frac{1}{2e(1 + \log x)^2} + C\].

Since the e factor was part of the exponentiation adjustment with the substitution, the final integral simplifies to:

\[-\frac{1}{2(1+\log x)^2} + c\].

This matches the provided correct answer option: \(-\frac{1}{2(1+\log\,x)^2}+c\).