 Hello friends and how are you all today? My name is Priyanka and I shall be helping you with the following question It says integrate the following rational functions. Now here the function which we need to integrate is 1 upon x plus 1 multiplied by x plus 2 now if we want to evaluate The integral of the function which is a proper rational function It's always possible to write the integrant as a sum of simpler rational functions by a method called partial fraction decomposition so we have the key idea towards this question as integration by partial fractions Now here We can write a proper rational function px plus q divided by x plus a into x plus b as The sum of simpler rational fractions that is a upon x plus a Plus b upon x plus b Right and in this the integration can be carried out easily by the already known methods and also Here we consider for integration purpose Only those denominators which can be factored into linear or quadratic factors Right, so this is the key idea of the question Proceeding on with the solution now here in this question We are given the rational function as x plus 1 into x plus 2 and we will take it equal to a upon x plus 1 plus b upon x plus 2 now taking LCM we have 1 over x plus 1 into x plus 2 is equal to a into x plus 2 Plus b into x plus 1 Divided by the LCM that was x plus 1 into x plus 2 right, so For this we can have 1 is equal to a x plus 2 plus b x plus 1 now we need to compare various terms like coefficients of power of x Constant to form linear equation Solving which we will get the values of a and b now comparing the coefficients of x on both the sides we have we have a plus b Equals to 0 Which implies the value of a is equal to minus b right and Comparing the constant terms on both sides We have a Plus b Equal to 1 now on substituting The value of a is equal to minus b in here We have to multiply by a which we have taken as minus b Plus b is equal to 1 which gives us Minus 2b plus b is equal to 1 which further implies minus b is equal to 1 and The value of b is equal to 1 now if the value of b is equal to is equal to Minus 1 Then we can say that the value of a is equal to a positive 1 right now on substituting the values of a and b we get 1 over x plus 1 into x plus 2 is equal to a Upon x plus 1 the value of a is 1 Minus 1 upon x plus 2 since the value of b was minus 1 therefore integrating both sides we get 1 over x plus 1 Into x plus 2 dx is equal to integral x plus 1 minus 1 over x plus 2 dx now taking integration sign separately with both the terms we get Integral of 1 over x plus 1 dx minus integral of 1 over x plus 2 dx and It is equal to log mod x plus 1 Minus log mod x plus 2 plus c and it can also be written as log x plus 1 upon x plus 2 Plus c so the required answer to the session is log mod of x plus 1 over x plus 2 Plus c right so this is the required answer to the session Hope you understood the whole concept of partial fractions well and have a nice day ahead