 Hello and welcome to the session. Let us discuss the following question. Question says, choose the correct answer. Integral dx upon x multiplied by x squared plus 1 equals log x minus 1 upon 2 log x squared plus 1 plus c log x plus 1 upon 2 log x squared plus 1 plus c c part is minus log x plus 1 upon 2 log x squared plus 1 plus c d part is 1 upon 2 log x plus log of x squared plus 1 plus c Let us now start with the solution. Now we have to find the integral dx upon x multiplied by x squared plus 1 Now first of all let us consider 1 upon x multiplied by x squared plus 1 Now we can write it as a upon x plus bx plus c upon x squared plus 1 Now this further implies 1 is equal to a multiplied by x squared plus 1 plus bx plus c multiplied by x Now we get 1 is equal to a x squared plus a plus bx squared plus cx Now comparing coefficients of x, x squared and constant terms on both the sides we get c is equal to 0, a is equal to 1 and a plus b is equal to 0 Now solving these three equations we get a is equal to 1, b is equal to minus 1 and c is equal to 0 Now substituting values of a, b and c in this expression we get 1 upon x multiplied by x squared plus 1 is equal to 1 upon x plus minus x upon x squared plus 1 Now integral of dx upon x multiplied by x squared plus 1 is equal to integral of 1 upon x dx plus integral of minus x dx upon x squared plus 1 Now let us name this integral as i1 and this integral as i2 First of all let us solve i1 You know i1 is equal to integral of 1 upon x dx Now this is further equal to log x plus c Here we can directly apply this formula of integration Now we know i2 is equal to this integral Now this can be further written as minus integral of x dx upon x squared plus 1 Now multiplying and dividing this integral by 2 we get minus 1 upon 2 multiplied by integral 2x dx upon x squared plus 1 Now put x squared plus 1 is equal to t Now differentiating both sides with respect to x we get 2x dx is equal to dt Now we get minus 1 upon 2 multiplied by integral dt upon t We know 2x dx is equal to dt and x squared plus 1 is equal to t Now this integral is further equal to minus 1 upon 2 log of t plus c Clearly we can see we can find this integral by using this formula of integration Here x has been replaced by t Now substituting x squared plus 1 for t we get minus 1 upon 2 log x squared plus 1 plus c Now we will substitute this value of i1 and this value of i2 In this expression Now we get integral of dx upon x multiplied by x squared plus 1 is equal to log x minus 1 upon 2 log of x squared plus 1 plus c Clearly we can see this is i1 and this is i2 So our correct answer is A This is our required answer This completes the session Hope you understood the solution Take care and keep smiling