 Hi and welcome to the session. Let us discuss the following question. Question says integrate the following rational functions. Given function is x square plus 1 multiplied by x square plus 2 upon x square plus 3 multiplied by x square plus 4. Let us now start with the solution. Now we are given a function x square plus 1 multiplied by x square plus 2 upon x square plus 3 multiplied by x square plus 4. Now this function is further equal to x raised to the power 4 plus 3x square plus 2 upon x raised to the power 4 plus 7x square plus 12. Multiplying these two brackets we get this expression and multiplying these two brackets we get this expression. Now clearly we can see this integrant is not a proper rational function as degree of numerator and denominator is same. So we will divide numerator by denominator. Let us now start with the division. Now we will multiply this divisor by 1 and we get x raised to the power 4 plus 7x square plus 12. Now subtracting the light terms we get minus 4x square minus 10. Now this integrant is equal to 1 plus minus 4x square minus 10 upon x raised to the power 4 plus 7x square plus 12. Now this is further equal to 1 plus minus 4x square minus 10 upon x square plus 3 multiplied by x square plus 4. On factorization of this polynomial we get these two vectors. Now consider this rational function and put x square is equal to t in this function. Now substituting t for x square we get minus 4t minus 10 upon t plus 3 multiplied by t plus 4. Now let us assume that this rational function is equal to a upon t plus 3 plus b upon t plus 4 so that minus 4t minus 10 is equal to a multiplied by t plus 4 plus b multiplied by t plus 3. Now equating the coefficients of t and constant terms on both sides we get minus 4 is equal to a plus b and minus 10 is equal to 4a plus 3b. Now let us name this equation as 1 and this equation as 2. Now from equation 1 we get a is equal to minus 4 minus b. Now substituting this value of a in equation 2 we get minus 10 is equal to 4 multiplied by minus 4 minus b plus 3b. Now solving this equation further we get minus 10 is equal to minus 16 minus 4b plus 3b. Multiplying 4 by this bracket we get minus 16 minus 4b and we have written plus 3b as it is. Now adding 16 on both the sides we get 6 is equal to minus 4b plus 3b. Now these are like terms minus 4b plus 3b is equal to minus b so we get 6 is equal to minus b. Now multiplying both the sides by minus 1 we get minus 6 is equal to b or we can simply write it as b is equal to minus 6. Now substituting value of b equal to minus 6 in equation 1 we get minus 4 is equal to a plus minus 6. Now multiplying these two signs we get a minus 6 is equal to minus 4. Now adding 6 on both the sides we get 2 is equal to a or we can simply write a is equal to 2. Now substituting values of a and b in this expression we get minus 4t minus 10 upon t plus 3 multiplied by t plus 4 is equal to 2 upon t plus 3 minus 6 upon t plus 4. Now given integrand in terms of t is t plus 1 multiplied by t plus 2 upon t plus 3 multiplied by t plus 4. Now this is equal to 1 plus 2 upon t plus 3 minus 6 upon t plus 4. Now substituting x square for t in this expression we get x square plus 1 multiplied by x square plus 2 upon x square plus 3 multiplied by x square plus 4 is equal to 1 plus 2 upon x square plus 3 minus 6 upon x square plus 4. We know x square is equal to t so we can replace t by x square in this expression. Now we will find out integral of given function with respect to x. So we will write integral of x square plus 1 multiplied by x square plus 2 upon x square plus 3 multiplied by x square plus 4 dx is equal to integral of 1 dx plus 2 multiplied by integral of dx upon x square plus 3 minus 6 multiplied by integral of dx upon x square plus 4. Now we know integral of 1 with respect to x is equal to x only. Now integral of this function is 2 multiplied by 1 upon root 3 tan inverse x upon root 3. And integral of this function is 6 multiplied by 1 upon 2 tan inverse x upon 2 plus c. Now let us understand how we have find integral of this function. We know formula of integration that integral of 1 upon x square plus a square dx is equal to 1 upon a tan inverse x upon a plus c. Now to find this integral we can write 3 as square of root 3. So using the formula of integration that integral of dx upon x square plus a square is equal to 1 upon a tan inverse x upon a plus c. Here a has been replaced by root 3. So we get this integral is equal to 1 upon 3 tan inverse x upon root 3 and we have written this 2 as it is. Now clearly we can see 4 can be written as square of 2. So again we will apply this formula of integration for finding this integral. And this integral is equal to 1 upon 2 multiplied by tan inverse x upon 2 and we have written this minus 6 as it is. Here c is the constant of integration. Now we will cancel common factor 2 from numerator and denominator both here. And we get integral of x square plus 1 multiplied by x square plus 2 upon x square plus 3 multiplied by x square plus 4 with respect to x is equal to x plus 2 upon root 3 tan inverse x upon root 3 minus 3 multiplied by tan inverse x upon 2 plus c. So this is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.