 Hello, and welcome to the session I am the speaker here. Let's discuss a question which says, integrate the problem rational function 3x plus 5 upon x cubed minus x squared minus x plus 1. That it is always possible to write the integral as a sum of severe rational functions by a method called partial friction decomposition. If the form of the rational function x squared plus qx plus r upon b whole square into x minus b, then form of the partial friction is a over x minus a plus x minus a whole square plus c over x minus b. There are real numbers to be determined suitably. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Integrate the rational function 3x plus 5 upon minus x squared minus x plus x plus 5 upon x plus 1 into x minus 1 whole square. The idea, or by using the form of partial fractions, vk is equal to to be determined. This gives us 5 whole 2 a into x minus 1 whole square plus b into x squared minus 1 plus c x plus 1. So 3x plus 5 is equal to x squared minus 2x plus 1 to x squared minus 1 plus c into x plus 1 is equal to a x squared minus a plus b x squared minus b plus the efficiency of x squared constant term is equal to 0. Let us get this equation as number 1 on equating the coefficient of x we get is equal to 3. Let us get this equation as number 2 and on equating the constant term we get a let us get this equation as number 3. Equation 1 in b is equal to minus a b is equal to minus a equation 3. We get is equal to 5. Let us get this equation as number 4. Equation 2 equation 4 equation 2 was an equation 4 was is equal to 5 is equal to 8. Therefore c is equal to 4. In equating the value of c equation 4 we get minus 2 plus 4 is equal to 3. This implies minus 2 a is equal to 3 minus 4. Again this implies minus 2 a is equal to minus 1 is equal to 1 by 2. Now equal to minus a implies is equal to minus 1 by 2 minus x squared minus x plus 1 is equal. So this is equal to 1 by 2 into x plus 1 minus 1 by 2 into x minus equal to 4. So 4 upon x minus 1 whole square. Therefore x plus 5 upon minus x square minus x plus 1 dx is equal to 1 by 2 into integral of 1 over x plus 1 by 2 into integral of 1 over x minus 1 dx 1 over x minus 1 whole square dx. The process of differentiation to 1 by 2 into integral of 1 over x plus 1 dx is equal to 1 by 2 log of x. So 1 over x minus 1 dx is equal to 1 by 2 log of 1 over x minus 1 whole square dx is equal to 4 into integral of 1 over x minus 1 whole square dx is equal to 4 is equal to minus 4 over x minus 1. Therefore s1 dx x minus 1 log of s1 plus c. 1 by 2 is clear to you by h.