 Hello friends and how are you all today? My name is Priyanka and the question says, integrate the following rational functions. Now here the function which is given to us is 1 upon x into x raised to the power n plus 1. Now when we are dealing with trigonometric functions it is always good to see if the numerator in the fraction can be obtained by differentiating a component in the denominator. We know that derivative of x raised to the power n is n into x raised to the power n minus 1. So multiplying the numerator denominator by n into x raised to the power n minus 1, we get 1 upon x into x raised to the power n plus 1 is equal to n into x raised to the power n minus 1 upon n into x raised to the power n minus 1 into x into x raised to the power n minus 1, sorry plus 1. Now integrating both the sides we get integral of 1 upon x into x raised to the power n plus 1 dx is equal to integral of n into x raised to the power n minus 1 upon n. Now x raised to the power n minus 1 when it gets multiplied by x we will have the answer as x raised to the power n into x raised to the power n plus 1 into dx. Now the numerator that is n into x raised to the power n minus 1 is clearly the derivative of the component x raised to the power n in the denominator. So put x raised to the power n equal to t and therefore differentiating both the sides we will have n into x raised to the power n minus 1 dx is equal to dt, right? So the above fraction will become 1 upon n into 1 upon t plus 1 dt. Now 1 upon n will be a constant so taking the constant out of the integral we have 1 upon n integral 1 upon t t plus 1 dt. Now 1 upon t t plus 1 can be solved by using partial fraction method. So we have now let us solve it by using partial fraction method separately. So we have 1 upon t t plus 1 can be written as a upon t plus b upon t plus 1. Now taking LCM we have and equating the numerator is equal to 1 is equal to a into t plus 1 plus b into t. So we have on equating the constants so it will be equal to a that is equal to 1 and on equating the coefficients of we have a plus b equal to 0 since the value of a is 1 so value of b will be equal to minus 1 isn't it? So we can write this as 1 upon n integral the value of a was obtained to be 1 so it will be 1 upon t plus the value of b was minus 1 upon t plus 1 into dt. Now taking integration sign separately with each term we have 1 upon n integral 1 upon t dt minus 1 upon n integral 1 upon t plus 1 dt that is equal to 1 upon n log mod t minus 1 upon n log t plus 1. So it can be also written as 1 upon n log t upon t plus 1 plus c. Now substituting the value of t as x raise to the power 1 we have 1 upon n log x raise to the power n divided by x raise to the power n plus 1 plus c. And this was the required answer to this session. Hope you enjoyed and understood the whole concept well. Have a nice day ahead.