 Hello and welcome to the session. In this session we will discuss about the integrals of some particular functions. Here we mentioned some important formulae of integrals and applied them for integrating many other related standard integrals. The first one goes like integral of dx upon x square minus a square is equal to 1 upon 2 a log modulus x minus a over x plus a plus c. Next is integral of dx upon a square minus x square is equal to 1 upon 2 a log modulus a plus x over a minus x plus c. Then the next is integral of dx upon x square plus a square is equal to 1 upon a tan inverse x upon a plus c. Then integral of dx upon square root of x square minus a square is equal to log modulus x plus square root of x square minus a square plus c. Then we have integral of dx upon square root of a square minus x square is equal to sin inverse x upon a plus c. Next one is integral of dx upon square root of x square plus a square is equal to log modulus x plus square root of x square plus a square plus c. We need to remember these integrals for integrating many other related standard integrals. Let's try and find out the value of the integral i equal to integral of dx upon x square minus 25. This can be written as integral of dx upon x square minus 5 square. Looking at the denominator we can very well see that we have to use the first formula which goes like integral of dx upon x square minus a square is equal to 1 upon 2 a log of modulus x minus a over x plus a plus c. So in this case we are going to take as 5 and we will substitute this value of a in the above formula so we get 1 upon 2 a log modulus x minus a over x plus a plus c. That is we get i is equal to 1 upon 10 log modulus x minus 5 over x plus 5 plus c. Here the c is the constant of integration. Next is if we have the integrals of the type integral dx upon a x square plus b x plus c integral of dx upon square root of a x square plus b x plus c integral of p x plus q dx upon a x square plus b x plus c and integral p x plus q dx upon square root of a x square plus b x plus c. Then we can reduce them to the form so that we can apply the above discussed formulae to find the value of the integrals. Let's try and find out the value of the integral i equal to integral of dx upon square root of x square plus 2 x plus 2. Let us try and reduce this integral to a form so that we can apply any of the above standard six formulas that we had discussed. For this what we do is we write x square plus 2 x plus 2 as x square plus 2 x plus 1 plus 1. This is further written as x plus 1 d whole square plus 1 square. That is now i is equal to integral of dx upon square root of x plus 1 d whole square plus 1 square. Then we substitute x plus 1 as t. This gives dx is equal to dt. So now this i becomes equal to integral of dt upon square root of t square plus 1 square. If you look at the denominator you can easily see that we can use the six standard formulae which is integral of dx upon square root of x square plus a square is equal to log of modulus x plus square root x square plus a square plus c. In the formula we will substitute x as t and a as 1. So we get this is equal to log of modulus x that is t plus square root of x square plus a square that is t square plus 1 square plus c. Now we substitute this value of t which is x plus 1. This gives us i is equal to log modulus x plus 1 plus square root of x plus 1 d whole square plus 1 plus c. That is equal to log modulus x plus 1 plus square root of x square plus 2x plus 2 plus c. This is the value for the integral i. Here the c is the constant of integration. In this way we can reduce the given integrals to the form so that we can apply the standard formulae that we had discussed. This completes the session. Hope you have understood the integration of some particular functions and how do we apply the standard formulas.