Correct Answer - Option 3 :
\(\tan \left( {1 + \log x} \right) + C\)
Concept:
-
\(\smallint {\sec ^2}x\;dx = \tan x + c\), where c is a constant
Integration by Substitution:
- If the given integration is of the form \(\smallint {\rm{g}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right){\rm{f'}}\left( {\rm{x}} \right){\rm{dx}}\) where \({\rm{g}}\left( {\rm{x}} \right)\) and \({\rm{f}}\left( {\rm{x}} \right)\) are both differentiable functions then we substitute \({\rm{f}}\left( {\rm{x}} \right) = {\rm{u}}\) which implies that \({\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{du}}\).
- Therefore, the integral becomes \(\smallint {\rm{g}}\left( {\rm{u}} \right){\rm{du}}\) which can be solved by general formulas.
Calculation:
Given: \(\smallint \frac{{dx}}{{x\; \times\; {{\cos }^2}\left( {1\; +\; \log x} \right)}}\)
Let 1 + log x = t
Now by differentiating the above equation with respect to t we get,
\(\Rightarrow \frac{{d\left( {1 + \log x} \right)}}{{dx}} = \frac{{dt}}{{dx}}\)
\(\Rightarrow \frac{1}{x} \cdot dx = dt\)
\(\Rightarrow \smallint \frac{{dx}}{{x{{\cos }^2}\left( {1 + \log x} \right)}} = \;\smallint \frac{{dt}}{{{{\cos }^2}t}} = \;\smallint {\sec ^2}t\;dt\)
As we know that, \(\smallint {\sec ^2}x\;dx = \tan x + c\) , where c is a constant
\(\Rightarrow \smallint {\sec ^2}t\;dt = \tan t + C\)
Now by substituting 1 + log x = t in the above equation we get,
\(\Rightarrow \smallint \frac{{dx}}{{x{{\cos }^2}\left( {1 + \log x} \right)}} = \tan \left( {1 + \log x} \right) + C\)
