 Hello and welcome to the session. In this session we will discuss the method for integration by parts. This method is basically very useful in integrating products of functions. For given functions f1 and f2 integral of the product of these two functions would be equal to the first function f1x into integral of the second function f2x minus integral of the differential coefficient of the first function multiplied by the integral of the second function. We have to be very careful in choosing the first function and the second function. The second function would be that function whose integral is well known to us and if any function is the power of x or a polynomial in x then we take that function as the first function and if in some cases if one function is inverse trigonometric function logarithmic function then these functions would be taken as the first function. Let's try and find out the value of the integral i equal to integral of x into e to the power x dx. Here the two functions are x and e to the power x. We know that if a function is a power of x or a polynomial in x then we take it as the first function. So in this case this x would be taken as the first function and e to the power x is taken as the second function. Now we integrate this by parts. On doing this we get the first function that is x into integral of the second function minus integral of the differential coefficient of the first function into integral of the second function. This gives us x into now integral of e to the power x is e to the power x itself minus integral of differential coefficient of x would be 1 into integral of e to the power x is e to the power x dx which further is equal to x into e to the power x minus e to the power x plus c. Remember that we do not add the constant of integration to the integral of the second function. So this is the final value for i. Here c is the constant of integration. If we have the integral of the type integral of e to the power x multiplied by fx plus f dash x dx this would be equal to e to the power x into fx plus c. Consider the integral i equal to integral of e to the power x into 1 upon x minus 1 upon x square dx. Now as you can see this integral is of this form integral of e to the power x into fx plus f dash x dx. Here we have fx is the function 1 upon x and its f dash x is equal to minus 1 upon x square. So from this we see that the value of this integral i is equal to e to the power x into fx that is 1 upon x plus c where the c is the constant of integration. Now we discuss some special types of integrals. The first one here is integral of square root x square minus a square dx is equal to x upon 2 square root x square minus a square minus a square upon 2 log modulus x plus square root x square minus a square plus c. The next one is integral square root of x square plus a square dx this is equal to x upon 2 square root x square plus a square plus a square upon 2 log modulus x plus square root x square plus a square plus c. Then integral square root a square minus x square dx is equal to x upon 2 square root a square minus x square plus a square upon 2 sin inverse x upon a plus c. Let's try and find the value of the integral i equal to integral of square root 4 minus x square dx. This is written in the form integral of square root 2 square minus x square dx. Now if you look at this function you can very well see that we can use this formula according to which we have integral of square root a square minus x square dx is equal to x upon 2 multiplied by square root a square minus x square plus a square upon 2 sin inverse x upon a plus c. In this case we will take a as 2 and we will substitute this value of a in this formula we get this is equal to x upon 2 square root of a square minus x square plus a square upon 2 sin inverse x upon a plus c. This is equal to x upon 2 square root of 4 minus x square plus 2 sin inverse x upon 2 plus c. This is the value for i. This c is the constant of integration. This is how we use this formulae to find the value of the integrants. This completes the session. Hope you understood the method of integration by parts.