 Hello friends and how are you all doing today? My name is Priyanka and the question says evaluate integral tan inverse x upon 1 plus x to the whole square tx. Let's quickly proceed with the solution. Let us rewrite the given integral once again. Tan inverse x upon 1 plus x to the whole square tx. First of all let us bring 1 plus x to the whole square in the numerator by making its exponent negative. Now here by applying by-parts we can write here we will take tan inverse x as the first function and this as the second function. So by applying by-parts we have tan inverse x into that is the first function into integral of second minus integral derivative of the first function into integral of the second function to dx. That is by applying by-parts. So on solving it out we have I equal to tan inverse x into 1 plus x minus 2 plus 1 upon minus 2 plus 1 minus integral. Derivative of tan inverse x is 1 upon 1 plus x square into again integral that is 1 plus x raised to the power minus 2 plus 1 upon minus 2 plus 1 into dx. Further I is equal to tan inverse x having minus sign upon 1 plus x minus integral dx upon this whole function will be 1 plus x and we have 1 plus x square already in the denominator. Now by using partial fraction we have minus tan inverse x upon 1 plus x minus Now here this function can be written as integral 1 by 2 integral dx upon 1 plus x that means a was found out by using partial fraction as 1 by 2 and therefore it is 1 by 2 integral 1 minus x upon 1 plus x square dx. Further we have I is equal to minus tan inverse x upon 1 plus x plus 1 by 2 this will be log mod 1 plus x plus an inverse x minus 1 by 2 log 1 plus x square plus c and this is the required answer to the given question. I hope you understood the whole concept well and enjoyed it too. Have a nice day ahead.