 Hello and welcome to the session. Let us discuss the following question. Question size. Choose the correct answer. The antiderivative of root x plus 1 upon root x equals a 1 upon 3 x raised to the power 1 upon 3 plus 2 multiplied by x raised to the power 1 upon 2 plus c b 2 upon 3 multiplied by x raised to the power 2 upon 3 plus 1 upon 2 multiplied by x square plus c c 2 upon 3 multiplied by x raised to the power 3 upon 2 plus 2 multiplied by x raised to the power 1 upon 2 plus c d 3 upon 2 multiplied by x raised to the power 3 upon 2 plus 1 upon 2 multiplied by x raised to the power 1 upon 2 plus c We have to choose the correct answer from a b c and d. First of all, let us understand that integral of fx plus gx dx is equal to integral of fx dx plus integral of gx dx and integral of x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1 plus c, where c is the constant of integration. Now we will use these two expressions as our key idea to solve the given question. Let us now start with the solution. Now we have to find integral of root x plus 1 upon root x dx. Now using expression 1 of key idea, we can write this integral as integral of root x dx plus integral of 1 upon root x dx. Now we know root x can be written as x raised to the power 1 upon 2. So here we will write integral of x raised to the power 1 upon 2 dx plus integral of x raised to the power minus 1 upon 2 dx. We know rule of exponents that 1 upon y raised to the power m is equal to y raised to the power minus m. So here it is 1 upon x raised to the power 1 upon 2. So we can write it as x raised to the power minus 1 upon 2. Now integral of x raised to the power 1 upon 2 dx is equal to x raised to the power 1 upon 2 plus 1 upon 1 upon 2 plus 1 plus integral of x raised to the power minus 1 upon 2 dx is equal to x raised to the power minus 1 upon 2 plus 1 upon minus 1 upon 2 plus 1 from expression 2 of the key idea we know integral of x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1. Here value of n is 1 upon 2 and here it is minus 1 upon 2. So, using the integral given in the key idea we will find integrals of these two functions. Here we will write plus c, c is the constant of integration. Now, adding these two terms we get 3 upon 2. So, we can write it as x raised to the power 3 upon 2 upon 3 upon 2 plus x raised to the power 1 upon 2. We know adding these two terms we get 1 upon 2 upon 1 upon 2 plus c. Now, this is further equal to 2 upon 3 multiplied by x raised to the power 3 upon 2 plus 2 upon 1 multiplied by x raised to the power 1 upon 2 plus c or we can further write it as 2 upon 3 x raised to the power 3 upon 2 plus 2 x raised to the power 1 upon 2 plus c. So, we can write antiderivative or we can say integral of root x plus 1 upon root x dx is equal to 2 upon 3 multiplied by x raised to the power 3 upon 2 plus 2 multiplied by x raised to the power 1 upon 2 plus c. So, the correct answer is c. c is the correct answer. This completes the session. Hope you understood the solution. Take care and have a nice day.