Question

Find the value of the integration \[\int\limits_1^{{e^2}} {\dfrac{{dx}}{{x{{\left( {1 + \log x} \right)}^2}}}} \] .
A.\[\dfrac{2}{3}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{3}{2}\]
D. \[\ln 2\]

Answer 1

Hint: First suppose that \[1 + \log x = t\], then differentiate both sides of the equation \[1 + \log x = t\] with respect to x, to obtain the value of \[dx\]. Then obtain the values of the upper and lower limit substitute 1 and \[{e^2}\] respectively in the equation \[1 + \log x = t\]. Then integrate the obtained equation to obtain the required result.

Formula used:
\[\int {\dfrac{1}{{{t^n}}}dt = \dfrac{{{t^{ - n + 1}}}}{{ - n + 1}}} \]

Complete step by step solution:
The given equation is,
\[\int\limits_1^{{e^2}} {\dfrac{{dx}}{{x{{\left( {1 + \log x} \right)}^2}}}} \].
Suppose that, \[1 + \log x = t\].
Now, differentiate both sides of the equation \[1 + \log x = t\]to obtain the value of dx.
\[\dfrac{{dx}}{x} = dt\]
Substitute 1 for x in the equation \[1 + \log x = t\] to obtain the value at x=1.
\[t = 1 + \log 1\]
\[t = 1 + 0\]
\[t = 1\]
Substitute \[{e^2}\] for x in the equation \[1 + \log x = t\] to obtain the value at \[x = {e^2}\].
\[t = 1 + \log {e^2}\]
\[t = 1 + 2{\log _e}e\]
\[t = 3\]
Therefore, the given equation becomes,
\[\int\limits_1^3 {\dfrac{{dt}}{{{t^2}}}} \]
\[ = \left[ {\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_1^3\]
\[ = \left[ { - \dfrac{1}{t}} \right]_1^3\]
\[ = - \dfrac{1}{3} + 1\]
\[ = \dfrac{2}{3}\]
The correct option is A.

Note: We use definite integrals when the upper and lower limits of that function are given. We use indefinite integrals when no limits are given to a particular function. Here we are using the definite integral. Students sometime do the calculation but did not change the limits, that leads to an incorrect solution. So, change the limit as well to calculate the integral correctly.