 Hello students, let's work out the following problem. It says evaluate the integral x upon x to the power 4 minus x square plus 1 So let's now start the solution and let I be the integral x upon x to the power 4 minus x square plus 1 dx. Now this can be written as integral x to the power 4 can be written as x square to the power 2 minus x square plus 1 into dx Now we know that the derivative of x square with respect to x is 2x. So we put x square is equal to t. So 2x into dx is equal to dt. So x dx is equal to dt by 2 So this integral becomes x dx is dt by 2. So We'll have 1 by 2 outside the integral. So dt upon t square minus t plus 1. Now we'll try to make this as a perfect square. So t square minus t plus 1. Now we add and subtract half the coefficient of t. We add and subtract the square of the half the coefficient of t which is half the coefficient of t is 1 by 2 and its square is 1 by 4. So we add and subtract 1 by 4. So we have 1 by 2 into integral dt upon t square plus 1 by 4 plus minus t is the whole square of t minus 1 by 2 and here we have 1 minus 1 by 4 and 1 by 2 into integral dt upon t minus 1 by 2 whole square 1 minus 1 by 4 is 3 by 4 and Again this can be written as 1 by 2 integral dt upon t minus 1 by 2 whole square 3 by 4 can be written as root 3 by 2 whole square and here we have plus 3 by 4. So now we know that the integral of 1 upon a square plus x square dx is given by 1 upon a into tan inverse x upon a plus c. You need to remember this formula. So now here x is t minus 1 by 2 and a is root 3 by 2 So this becomes 1 by 2 into 1 upon a that is 1 upon root 3 by 2 tan inverse t minus 1 by 2 upon root 3 by 2 plus c So again this is equal to 1 by 2 into 2 by root 3 tan inverse 2t minus 1 upon 2 upon root 3 by 2 plus c 2 gets cancelled with 2 and here also. So we are left with 1 by root 3 tan inverse 2t minus 1 upon under the root 3 plus c. Now we know that t is x square. So we will substitute it. So we have 1 upon root 3 tan inverse 2 into x square minus 1 upon root 3 plus c and this is the required value of the integral. So this completes the question and the session. Bye for now. Take care and have a good day.