 Hi, and how are you all today the question says integrate the following rational functions I had a function which is given to us is 1 upon x square minus 9 So here we will be applying the concept of integration by partial fractions that is px plus q upon x plus a into x plus b can be written as a upon x plus a plus b upon x plus b and This is the key idea of The question if you notice the rational function which is given to us can be written as one over x plus 3 into x plus sorry x minus 3, right? So now here we will we can use the concept of Partial fractions to find out the integration of this function So let us proceed on we can write the function as 1 over x plus 3 into x minus 3 now equal to a upon x plus 3 Plus b upon x minus 3, right? Now taking else him on both the sides we have 1 upon x plus 3 into x minus 3 equal to a into x minus 3 plus b into x plus 3 divided by x plus 3 into x minus 3 now Comparing The coefficients of x and then the constants we have the two equations as a plus b equal to 1 Which gives us the value of a as minus b? Let this be the first equation and then by comparing the coefficient that is the constants we have minus 3 a plus plus 3 b equal to 1 Let this be the second equation now on substituting the value of a From the first equation in the second equation We have Minus 3 the value of a is minus b Plus 3 b is equal to 1 which is further equal to 3 b plus 3 b is equal to 1 That further implies the value of b is equal to 1 by 6 Now if the value of b is 1 by 6 therefore the value of a will be equal to minus 1 by 6 right now We have to substitute the value of a and b above so we have x plus 3 into x minus 3 is equal to the value of a is minus 1 by 6 and in the denominator we had x plus 3 Plus 1 over 6 x minus 3 now integrating both sides we have integral of 1 upon x plus 3 into x minus 3 into dx equal to integral of minus 1 upon 6 into x plus 3 Plus 1 upon 6 into x minus 3 Now taking integration sign separately with both the terms we get Now we can write this function as 1 upon x square minus 9 into dx is equal to minus 1 over 6 Integral 1 upon x plus 3 into dx Plus 1 over 6 integral of 1 upon x minus 3 dx That is equal to minus 1 upon 6 log mod of x plus 3 Plus 1 over 6 mod of x minus 3 plus c So on rearranging we can write it as 1 upon 6 Here we had lot of log also log of mod x minus 3 minus 1 over 6 log x plus 3 plus c Which can be written as? 1 over 6 taking common log mod of x minus 3 upon x plus 3 Plus c so the required answer to the session is 1 over 6 log mod x minus 3 upon x plus 3 Plus c wait this completes the session. Hope you understood it well. Have a nice day ahead