 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says integrate the following function. There is tan inverse x. So let us find the solution to this question. Let i be equal to integral tan inverse x dx. We can write this as integral tan inverse x into 1 dx. Now we see that this is a product of two functions. So what we do is this becomes a first function. This becomes a second function. Since we do not have any function along with tan inverse x, so we can say that this will be the first one. This will be the second one. That will be equal to tan inverse x that is first function into integral of second function that is 1 dx minus integral of d by dx of tan inverse x into integral of 1 dx into dx. This will be equal to tan inverse x into x minus integral 1 upon 1 plus x square into x dx. This is equal to x tan inverse x minus 1 by 2 to integral 2x by 1 plus x square dx. What we have done is we have multiplied and the numerator and denominator by 2 and we have taken 1 by 2 out of the integral sign. This will be equal to x tan inverse x minus. Now what we do here is we put 1 plus x square to be equal to t. So this implies that 2x dx will be equal to dt. So this we have minus 1 by 2 integral 2x dx is dt divided by t. This will be equal to x tan inverse x minus 1 by 2 integral dt by t is log t plus some constant c. Now we put back the value of t as 1 plus x square that is equal to x tan inverse x minus 1 by 2 log of mod 1 plus x square plus c. We have taken mod because we see that log cannot take negative values or we can write it like this only. So our answer to this question is x tan inverse x minus 1 by 2 log of 1 plus x square plus c. So I hope that you understood the question and enjoyed the session. Have a good day.