 Hi and welcome to the session. Let us discuss the following question. Question says, integrate the following functions. Given function is 1 upon square root of x square plus 2x plus 2. First of all, let us understand that integral dx upon square root of x square plus a square is equal to log of x plus square root of x square plus a square plus c where c is the constant of integration. We will use this formula as our key idea to solve the given question. Let us now start with the solution. We have to find integral 1 upon square root of x square plus 2x plus 2 dx. Now let us consider the expression x square plus 2x plus 2. Now it can be written as x square plus 2x plus 1 plus 1. 2 can be written as 1 plus 1. Now these three terms form whole square of x plus 1 and 1 can be written as it is. We know a square plus b square plus 2ab is equal to a plus b whole square. Now this is further equal to whole square of x plus 1 plus 1 square. We know 1 raised to any power is equal to 1 only. So we can write 1 as 1 square. So we get x square plus 2x plus 2 is equal to x plus 1 whole square plus 1 square. Now integral dx upon square root of x square plus 2x plus 2 is equal to integral dx upon square root of x plus 1 whole square plus 1 square. From this expression we get x square plus 2x plus 2 is equal to x plus 1 whole square plus 1 square. Now we know derivative of x plus 1 is equal to 1. So we will substitute x plus 1 equal to t. This implies 1 is equal to dt upon dx. Differentiating both the sides of this expression with respect to x we get 1 is equal to dt upon dx. Now this further implies dx is equal to dt. Now we can write integral dx upon square root of x plus 1 whole square plus 1 square is equal to integral of dt upon square root of t square plus 1 square. We know we can substitute t for x plus 1 and dt for dx. Now using formula given in key idea we get this integral is equal to log of t plus square root of t square plus 1 square plus c. Comparing the integrals given in key idea clearly we can see variable x has been replaced by t here and a is equal to 1. So we get log t plus square root of t square plus 1 square plus c. Now we know t is equal to x plus 1 so we will substitute x plus 1 for t. Now we get log of x plus 1 plus square root of x plus 1 whole square plus 1 we know square of 1 is equal to 1 only plus c. This is further equal to log of x plus 1 plus we know this expression is equal to x square plus 2x plus 2 plus c. Now our required answer is integral of dx upon square root of x square plus 2x plus 2 is equal to log of x plus 1 plus square root of x square plus 2x plus 2 plus c. This completes the session hope you understood the solution take care and have a nice day.