 Hello and welcome to the session. The question says integrate the following function and the function is 4x plus 1 upon root over 2x square plus x minus 3. So let's start with the solution and we have to integrate the given function so we have integral 4x plus 1 upon root over 2x square plus x minus 3 into dx. Now let us put t is equal to 2x square plus x minus 3. So differentiating both sides with respect to x on the left hand side we have dt upon dx and on the right hand side we have 4x plus 1. So dt is equal to 4x plus 1 into dx. So this integral can further be written as integral dt upon root over t which is further equal to integral t raised to the power minus half into dt. This is further equal to t raised to the power minus half plus 1 upon minus half plus 1 plus c since integral of the function x raised to the power n with respect to x is equal to x raised to the power n plus 1 upon n plus 1 plus c. Thus we have root over t upon 1 by 2 plus c or we have 2 root t plus c and t is square plus x minus 3 thus on integrating the given function we get the answer as 2 into 2x square plus x minus 3 plus c. So this completes the session by intake care.