 Hello and welcome to the session, the question says integrate the function 1 upon root over 9 minus 25 x square. So first let us learn the formula to integrate the function of the type 1 upon root over a square minus x square with respect to x. So this is given by sin inverse x upon a plus c. So with the help of this formula we shall integrate the given function. So this is our key idea. Now let us start with the solution. The given function is 1 upon root over 9 minus 25 x square. This can further be written as 1 upon root over 3 square minus 5 x whole square. Now we have to integrate this function. Therefore we have integral dx upon root over 3 square minus 5 x whole square. Let us put 5 x is equal to t. So this implies 5 is equal to dt upon dx integrating both sides with respect to x which further implies that dx is equal to dt upon 5. Thus this integral can further be written as integral dt upon 5 into root over 3 square minus t square. Taking 1 upon 5 outside the integral we have 1 upon 5 integral dt upon root over 3 square minus t square. Now let us apply the formula here in place of x we have t and in place of a we have 3. Therefore it can further be written as 1 upon 5 sin inverse x is t and a is 3 plus c where c is a constant. This is further equal to 1 upon 5 sin inverse t is 5 x upon 3 plus c. Thus on integrating the given function we get 1 upon 5 sin inverse 5 x upon 3 plus c. So this completes the session. Bye and take care.