 Hi and welcome to our session. Let us discuss the following question. The question says, integrate the following functions. The function is 1 by x minus x cube. Let us now begin with this illusion. In this question we have to find integral of 1 by x minus x cube with respect to x. Now take x cube common from the denominator. So we have 1 by x cube into 1 by x square minus 1 dx. Now you can see that 1 by x cube is the derivative of 1 by x square. So we will now put 1 by x square minus 1 as t. Now this implies dt by dx is equal to minus 2 by x cube. This implies minus 1 by 2 into dt is equal to dx by x cube. Now we will substitute t in place of 1 by x square minus 1 and minus 1 by 2 dt in place of dx by x cube. So this is equal to minus 1 by 2 into integral of dt by, we know that integral of dx by x is log x plus c. So this means integral of dt by t is log t plus c. Now t is equal to 1 by x square minus 1. So this is equal to minus 1 by 2 log 1 by x square minus 1 plus c. This is equal to minus 1 by 2 log 1 minus x square by x square plus c. This is equal to 1 by 2 log x square by 1 minus x square plus c. And the required answer is 1 by 2 log x square by 1 minus x square plus c. So this completes the session. Bye and take care.