 Hello and welcome to the session. Let us discuss the following question. Question says, integrate the functions. Given function is 2x upon 1 plus x square. First of all, let us understand that while doing integration by substitution method, we make a substitution for a function whose derivative also occurs in the integrand. This is the key idea to solve the given question. Let us now start with the solution. Now we have to integrate 2x upon 1 plus x square with respect to x. Now if we differentiate 1 plus x square with respect to x we get derivative of 1 plus x square is equal to 2x. We know derivative of 1 is 0 and derivative of x square is 2x. Now derivative of this function is present in numerator. Now using key idea, we make the substitution 1 plus x square is equal to t. Now differentiating both the sides with respect to x we get 2x dx is equal to dt. Now we can write integral of 2x upon 1 plus x square dx is equal to integral of dt upon t. Clearly we can see 2x dx is equal to dt. So here we will substitute dt for 2x dx and 1 plus x square is equal to t. So we will substitute t for 1 plus x square. Now integral of 1 upon t with respect to t is equal to log t plus c where c is the constant of integration. We know t is equal to 1 plus x square. So here we will substitute 1 plus x square for t. So we get integral of 2x upon 1 plus x square with respect to x is equal to log of 1 plus x square plus c. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.