 Hello students let's solve the following integral problem it says integrate the following function the given function is log x whole square upon x now to integrate this we first need to know the formula for the integral of y to the power n dy which is y to the power n plus 1 upon n plus 1 plus c where c is the constant of the integral and this will be the key idea let us now move on to the solution we have to find the integral of log x whole square upon x now here we see that the derivative of log x is 1 upon x so we substitute log x as y so put y is equal to log x now differentiating this with respect to x so dy by dx is equal to 1 upon x d so this implies dy is equal to 1 upon x dx now 1 upon x into dx is equal to dy and log x is y so the integral becomes y square dy now we integrate this and by the formula it is y to the power n plus 1 upon n plus 1 plus c now here n is 2 so this becomes y to the power 2 plus 1 that is 3 upon 3 plus c now we substitute the value of y so the value of y is log x so this becomes log x to the power 3 upon 3 plus c hence the integral of the given function is 1 upon 3 into log x cube plus c and this completes the question bye for now take care have a good day