 Hello and welcome to the session. The given question says evaluate the following definite integral, 15th is integral 2x plus 3 upon 5x square plus 1 dx and the lower limit of the integration is 0 and the upper limit of integration is 1. So let's start with the solution and first let us find the value of integral 2x plus 3 upon 5x square plus 1 with respect to x. So it can be written as integral 2 hangs of x upon 5x square plus 1 into dx plus 3 hangs integral dx upon 5x square plus 1. Now let us name this integral as i1 and this as i2. Now let us solve both these integrals one by one and then we will substitute the values here. Now the first integral i1 as integral x upon 5x square plus 1 into dx. So let us suppose that 5x square plus 1 is equal to t. So this implies 10x into dx is equal to dt or x into dx is equal to dt upon 10. So this integral can further be written as dt upon 10 and the value of 5x square plus 1 is t. Now taking the constant outside the integral sign we have dt upon t and this is equal to 1 upon 10 log mod t and the value of t is 5x square plus 1. So we further have 1 upon 10 log mod 5x square plus 1. So let us substitute the value here. We have 2 into 1 by 10 log mod 5x square plus 1. Now let us solve the second integral. It is i2 which is integral dx upon 5x square plus 1. It can further be written as integral dx upon root over 5x whole square plus 1 square. And now let us put 5 into x is equal to t. So this implies root 5 into dx is equal to dt or dx is equal to dt upon root 5. So here we have integral dt taking 1 by root 5 outside the integral sign and in the denominator we have t square plus 1 square and this is further equal to 1 upon root 5 tan inverse t since integral 1 upon 1 plus x square with respect to x is equal to tan inverse x. So this is further equal to 1 upon root 5 tan inverse t is root 5 into x. So let us substitute the value of i2 here. 3 we already have and the value of i2 is 1 upon root 5 tan inverse root 5 into x. So this is further equal to 1 by 5 log mod 5x square plus 1 plus 3 upon root 5 tan inverse root 5 into x. Now by the second fundamental theorem of integral calculus the value of the definite integral which is integral as 3 upon 5x square plus 1 into dx from 0 to 1 and the written as 1 upon 5 log mod 5x square plus 1 plus 3 upon root 5 tan inverse root 5x lower limit of the integral is 0 and the upper limit is 1. Since by the second fundamental theorem it says if we have integral fx dx and the lower limit of the integration is a and upper limit is b then it is equal to the value of the antiderivative of the given function at the point b minus the value of the antiderivative at the point a where f is a continuous function defined on the closed interval a to b. Now let us simplify this we have 1 upon 5 log mod 5x square plus 1 lower limit 0 and upper limit 1 plus 3 upon root 5 tan inverse root 5 into x from 0 to 1 and this is further equal to 1 by 5. Now putting upper limit in this function we have log mod 5 plus 1 minus log mod 1 plus 3 upon root 5 tan inverse root 5 minus tan inverse root 5 into 0 is 0. This is further equal to 1 by 5 log 6 minus log 1 plus 3 upon root 5 tan inverse root 5 minus tan 0 is 0 so tan inverse 0 is again 0 and this is further equal to 1 by 5 log 6 since log 1 is 0 plus 3 upon root 5 tan inverse root 5. Thus on evaluating the given integral our answer is 1 by 5 log 6 plus 3 upon root 5 tan inverse root 5. This completes the session by intake care.