 Hello and welcome to the session. In this session we discussed the following question it says evaluate integral 0 to 1 dx over 1 plus x square. First let's discuss the second fundamental theorem of integral calculus. It says that let f be a continuous function defined on the closed interval a, b and f, b and an nt derivative of f then integral a to b fx dx is equal to fx a to b that is equal to fb minus fa. This is the key idea for this question. Now let's move on to the solution. We take let i be equal to integral 0 to 1 dx over 1 plus x square. Now since integral dx over 1 plus x square is equal to tan inverse x let this be equal to fx. Therefore by second fundamental theorem we get i is equal to f of 1 minus f of 0. Now we have fx equal to tan inverse x so f of 1 would be equal to tan inverse 1 minus f of 0 that is tan inverse 0. Now tan inverse 1 is equal to pi by 4 minus tan inverse 0 that is 0 so we get i equal to pi by 4. Thus our final answer is integral 0 to 1 dx over 1 plus x square is equal to pi by 4. This completes the session. Hope you have understood the solution for this question.