 Hello and welcome to the session the question says integrate the following function and the given function is 1 upon root over 1 plus 4 x square. So first let us learn a simple integral formula with the help of which we shall integrate the given function. So if we have a function of the form 1 upon root over x square plus a square then let's integral with respect to x is equal to log x plus root over x square plus a square plus c. So with the help of this formula we shall integrate the given function so this is our key idea. Let's start with the solution. The given function is 1 upon root over 1 plus 4 x square which can further be written as 1 upon root over 2 x whole square plus 1 square. Now we have 1 upon root over x square plus 1 square into dx. Now let us put is equal to t. Now on integrating both sides with respect to x we have 2 into 1 is equal to dt upon dx or we have dx is equal to dt upon t. So the integral can further be written as dx is dt upon 2 inside the root we have 2x is equal to t so t square plus 1 square. Taking 1 upon 2 outside the integral sign we are left with dx upon root over t square plus 1 square. Now applying the formula which is our key idea we have half log t plus root over t square plus 1 square plus c we are seeing constant. Now putting t is equal to 2x we have 1 upon 2 log 2x plus root over 2x whole square plus 1 plus c or this is further equal to half log 2x plus root over 4x square plus 1 plus c. Thus on integrating the given function our answer is half log 2x plus root over 4x square plus 1 plus c. So this completes the presentation bye and take care.