 Hello and welcome to the session, the given question says evaluate the following definite integral, twelfth as integral dx upon 1 plus x square and the lower limit of integration is 0 and the upper limit of integration is 1. So, let us now start with the solution. First, we shall find the value of integral dx upon 1 plus x square. So, this is equal to tan inverse x. Now, by the fundamental theorem of integral calculus we know that integral f x dot dx from a to b is equal to f p minus f a where f is the anti-derivative of the given function f which is continuous and defined on interval a to b which is closed. So, from here we find that f x which is the anti-derivative of the given function is tan inverse x. Therefore, integral dx upon 1 plus x square from 0 to 1 is equal to and inverse x from 0 to 1 which is by the second fundamental theorem of integral calculus. So, this is equal to tan inverse 1 minus tan inverse 0. Now, tan pi by 4 is 1. So, tan inverse 1 is pi by 4 minus tan 0 is 0. So, here we have tan inverse 0 is equal to 0. So, this gives pi by 4. Thus, on evaluating the given integral we get the answer as pi by 4. So, this completes the session. Bye and take care.