 Hello and welcome to your session. My name is Kanika and I'm going to help you to solve the following question. The question says, integrate the following functions, given function as 1 by 1 plus x to the power 4. Let's now begin with the solution. In this question, we have to integrate the function 1 by 1 plus x to the power 4 with respect to x. Now, first of all, you should know that if we have to integrate the function of the form 1 by x to the power 4 plus lambda x squared plus 1. Now, if we have to integrate functions of the type 1 by x to the power 4 plus lambda x squared plus 1 dx, then in the first step, we divide numerator and denominator by x squared. Then in the second step, we try to express denominator as x plus 1 by x whole squared plus k squared or x minus 1 by x whole squared minus k squared. Then in the third step, we introduce derivative of x plus 1 by x in the numerator if we have x plus 1 by x in the denominator plus 1 by x if we have x minus 1 by x in the denominator. So, we have x plus p or x minus 1 by x as p as required by the question and then we evaluate the integral. So, keeping this in mind, let's now integrate the given function. First divide numerator and denominator by x squared. So, we have 1 by x squared plus x squared dA and divide this expression by 2. So, we have to integral of we can write 1 minus 1 by x. So, using is equal to integral of 1 plus 1 by x squared squared dx minus 1 by 2 into integral of 1 minus 1 by x squared x squared plus x squared denominator in the form of x minus 1 by acting 2 we get 1 by x squared plus x squared minus 2 plus 2. Now 1 by x squared plus x squared minus 2 is equal to x minus 1 by x whole squared plus 2 can be written as squared in the numerator plus 1 by x whole squared. Using this, now these integrals is equal to integral 1 by x as v in second integral. Now u equals to x minus 1 by x implies du is equal to 1 plus 1 by x squared dx v equals to x plus 1 by x implies dv is equal to 1 minus 1 by x squared dA substituting u in place of x minus 1 by x in the first integral we get into integral of 1 by u squared with respect to u and by substituting v in place of x plus 1 by x in second integral we get integral dv by v squared minus root 2 squared. We know that integral of with respect to x is 1 by a tan inverse plus c and integral of 1 by x plus a plus c. So using these formulas integral of u squared plus root 2 square is 1 by root 2 tan inverse minus integral of dv by v squared minus root 2 square is 1 by 2v minus root 2 by v plus root 2 plus c plus c is denoting the sum of both the constants of these integrals. We have assumed 1 by x as v. So now putting x minus 1 by x in place of u and x plus 1 by x in place of v we get 1 by minus root 2