 Hi, and welcome to the session. Let us discuss the following question. Question says, integrate the following functions. Given function is 5x minus 2 upon 1 plus 2x plus 3x squared. First of all, let us understand that integral dx upon x squared plus a squared is equal to 1 upon a tan inverse x upon a plus c where c is the constant of integration and integral dx upon x is equal to log x plus c where c is the constant of integration. Now these two formulas of integration are key idea to solve the given question. Let us start with the solution. Now we have to find the integral 5x minus 2 dx upon 1 plus 2x plus 3x squared. Where we can see this integral is of the form px plus q upon a x squared plus bx plus c. So we will put px plus q that is 5x minus 2 equal to a multiplied by derivative of 1 plus 2x plus 3x squared that is denominator of the given integral plus b Now we know derivative of 1 plus 2x plus 3x squared is equal to 2 plus 6x so we can write it as a multiplied by 2 plus 6x plus b is equal to 5x minus 2. Now comparing the coefficient of x on both the sides we get 6x is equal to 5 and this implies a is equal to 5 upon 6 Now comparing the constant terms on both the sides we get 2a plus b is equal to minus 2. Now substituting value as a 5 upon 6 in this expression we get 2 multiplied by 5 upon 6 plus b is equal to minus 2. Now we will cancel common factor 2 and we get b is equal to minus 2 minus 5 upon 3 Now this further implies b is equal to minus 11 upon 3. For practicing these two terms by taking their LCN we get minus 11 upon 3 So we can write 5x minus 2 is equal to 5 upon 6 multiplied by 2 plus 6x plus minus 11 upon 3 Now we get integral 5x minus 2 dx upon 1 plus 2x plus 3x square is equal to 5 upon 6 multiplied by integral 6x plus 2 dx upon 1 plus 2x plus 3x square plus minus 11 upon 3 multiplied by integral dx upon 1 plus 2x plus 3x square. Now let us name this integral as i1 and this integral as i2. So we can write this expression is equal to 5 upon 6 i1 minus 11 upon 3 i2. Now let us name this expression as 1 Now first of all we will find out integral i1 that is 6x plus 2 dx upon 1 plus 2x plus 3x square. Not really we can see 6x plus 2 is derivative of 1 plus 2x plus 3x square. So to find this integral we will use substitution method. So we can write put 1 plus 2x plus 3x square equal to t. Differentiating both sides with respect to x we get 2 plus 6x is equal to dt upon dx Now this can be further written as 6x plus 2 dx is equal to dt. Now substituting these two values in this integral we get integral 6x plus 2 dx upon 1 plus 2x plus 3x square is equal to integral dt upon t. Now using the second formula given in key idea we get dt upon t is equal to log t plus c where c is the constant of integration we also know that t is equal to 1 plus 2x plus 3x square. So substituting 1 plus 2x plus 3x square for t we get log of 1 plus 2x plus 3x square plus c. So we get integral i1 is equal to log of 1 plus 2x plus 3x square plus c. Now we will find the integral i2. Now we felt finding the integral i2. First of all let us consider the expression 1 plus 2x plus 3x square. Multiplying and dividing this expression by 3 we get 3 multiplied by 1 upon 3 plus 2 upon 3x plus x square. Now this can be further written as 3 multiplied by 1 upon 3 plus 2 multiplied by 1 upon 3 multiplied by x plus x square plus square of 1 upon 3 minus square of 1 upon 3. We know 2 upon 3x can be written as 2 multiplied by 1 upon 3 multiplied by x and here we have done addition and subtraction of the same term that is square as 1 upon 3. Now this expression is further equal to 3 multiplied by x plus 1 upon 3 whole square plus 1 upon 3 minus 1 upon 9. Now a square plus v square plus 2ab is equal to a plus b whole square and 1 upon 3 is written as it is and square of 1 upon 3 is equal to 1 upon 9. So we can write 1 upon 3 minus 1 upon 9 here. Now this is further equal to 3 multiplied by square of x plus 1 upon 3 plus 2 upon 9. Let us now find the integral i2 that is integral dx upon 1 plus 2x plus 3x square. Now we have shown about that 1 plus 2x plus 3x square can be written as 3 multiplied by whole square of x plus 1 upon 3 plus 2 upon 9. So this integral is equal to dx upon 3 multiplied by square of x plus 1 upon 3 plus 2 upon 9. Now this is further equal to 1 upon 3 integral dx upon square of x plus 1 upon 3 plus 2 upon 9. Now we know 2 upon 9 is square of root 2 upon 3. So we can write this integral as 1 upon 3 multiplied by integral dx upon x plus 1 upon 3 whole square plus root 2 upon 3 square where we can see derivative of x plus 1 upon 3 is 1. So we will solve this integral by substitution method. So we can write x plus 1 upon 3 is equal to 2. Now differentiating both sides with respect to x we get 1 is equal to dt upon dx. Now this further implies dx is equal to dt. Now let us find out this integral. We know this integral is equal to 1 upon 3 multiplied by dt upon p square plus square of root 2 upon 3 substituting dt for dx in this integral and t for x plus 1 upon 3 in this integral we get this integral is equal to this integral. Now using first formula given in key idea we get this integral is equal to 1 upon 3 multiplied by 3 upon root 2 multiplied by tan inverse t upon root 2 upon 3 plus c where c is the constant of integration. Now integral of dx upon x square plus a square is equal to 1 upon a multiplied by tan inverse x upon a plus c where c is the constant of integration. Here value as a is root 2 upon 3 and x has been replaced by variable t. Now simplifying this expression further and substituting value of t here we get 1 upon root 2 tan inverse x plus 1 upon 3 upon root 2 upon 3 plus c now this is further equal to 1 upon root 2 tan inverse 3x plus 1 upon root 2 plus c. So integral i2 is equal to 1 upon root 2 tan inverse 3x plus 1 upon root 2 plus c. Now we will substitute values of i1 and i2 in expression 1. Expression 1 is 5 upon 6 i1 minus 11 upon 3 i2. Now you know i1 is equal to log of 1 plus 2x plus 3x square. So we can write it as 5 upon 6 multiplied by log of 1 plus 2x plus 3x square minus 11 upon 3 multiplied by value of i2 is this. So here we will write 1 upon root 2 multiplied by tan inverse of 3x plus 1 upon root 2 plus c. Here c is the constant of integration Now this can be further written as 5 upon 6 log of 3x square plus 2x plus 1 minus 11 upon 3 root 2 tan inverse of 3x plus 1 upon root 2 plus c. So we get integral 5x minus 2 dx upon 1 plus 2x plus 3x square is equal to 5 upon 6 multiplied by log of 3x square plus 2x plus 1 minus 11 upon 3 multiplied by root 2 multiplied by tan inverse of 3x plus 1 upon root 2 plus c. So this is our required answer. This completes the session. If you understood the solution, take care and have a nice day.