 Hello and welcome to the session. In this session, we discuss the following question which says evaluate integral of 1 minus x square upon x into 1 minus 2x the whole dx. Before we go to the solution, let's see that integral of 1 upon x dx is equal to log of modulus x plus c. This is the key idea that we use in this question. Let's move on to the solution now. Let us suppose the given integral v equal to i that is i is equal to the integral 1 minus x square upon x into 1 minus 2x the whole dx. So i is equal to integral of 1 minus x square upon x minus 2x square dx. Now multiplying dividing the numerator of the integrand by 2 we get i is equal to integral of 1 upon 2 into 2 minus 2x square this whole upon x minus 2x square dx. So further i is equal to taking this constant 1 upon 2 outside the integral we get 1 upon 2 into integral 2 minus 2x square upon x minus 2x square dx. Now we can write this 2 minus 2x square minus x plus x minus 2x square thus is equal to 1 upon 2 integral 2 minus x plus x minus 2x square this whole upon x minus 2x square dx. We now have i is equal to 1 upon 2 into integral 2 minus x 2x square dx plus 1 upon 2 into integral x minus 2x square upon x minus 2x square dx. Now this x minus 2x square cancels with x minus 2x square and we have i is equal to 1 upon 2 into integral taking x common in this denominator x into 1 minus 2x the whole into dx plus 1 upon 2 into integral 1 dx. So we write this as i is equal to 1 upon 2 i1 we take this integral as i1 plus 1 upon 2 into integral 1 into dx. Now let us evaluate the integral i1 equal to integral 2 minus x upon x into 1 minus 2x the whole dx integration by partial fractions so we can write 2 minus x upon x into 1 minus 2x plus b upon 1 minus to find the values of a and b. So now we have 2 minus x upon x into 1 minus 2x the whole is equal to a into 1 minus 2x the whole plus b into x x into 1 minus 2x the whole. This means we have 2 minus x is equal to plus b minus 2a the whole into x by creating the constant terms we have a is equal to 2 and then equating the coefficients of x we get b minus 2a is equal to minus 1. Now putting the value of a as 2 in this we get b minus 4 is equal to minus 1 or you can say b is equal to 3. So that we have a equal to 2 and b equal to 3 and so therefore we can write upon x into 1 minus 2x the whole is equal to 2 upon x that is in place of a we put 2 plus 3 upon 1 minus 2x that is in place of b we put 3. Thus we can now write I1 equal to integral 2 upon x plus 3 upon 1 minus 2x the whole dx so this is equal to 2 into integral 1 upon x dx plus 3 into integral 1 upon 1 minus 2x dx. So this is the integral I1 now we already know that integral of 1 upon x dx is equal to log of modulus x plus c so this gives us I1 is equal to 2 into log of modulus x plus the coefficient of x which is minus 2 into log of modulus of 1 minus 2x constant of integration which is c. So we have I1 is equal to 2 into log of modulus x minus 3 by 2 into log of modulus of 1 minus 2x plus the constant of integration c. Now let us assume this equation as equation 1 the value of I1 in equation 1 I is equal to half into I1 that is half into 2 into log of modulus x minus 3 by 2 into log of modulus 1 minus 2x plus the constant of integration c half into integral of 1 into dx so this gives us I is equal to half into 2 log of modulus x minus 3 by 2 into log of modulus 1 minus 2x plus the constant of integration c the whole half into the constant of integration c1. This means we now have I is equal to log of modulus x minus 3 by 4 into log of modulus 1 minus 2x plus 1 upon 2c plus 1 upon 2x plus c1 is equal to log of modulus x minus 3 by 4 into log of modulus c. Modulus 1 minus 2x plus 1 upon 2 into x plus the constant of integration c2 where we have c2 is equal to c1 plus half therefore upon x into 1 minus 2x the whole dx is equal to log of modulus x minus 3 by 4 into log of modulus x minus 3 by 4 into log of modulus x minus 3 by 4 into log of modulus c. log of modulus 1 minus 2x 1 upon 2x plus c2 where the c2 is the constant of integration so this is our hope you understood the solution of this question.