 Hello friends. Let's solve the following question. It says integrate the following function The given function is root ax plus b. Let's now move on to the solution Let i be the integral root ax plus b dx now What t is equal to? to ax plus b. So dT by dx is equal to a and this implies dt is equal to adx and this again implies dx is equal to dt by a Now dx is dt by a and t is ax plus b. So substitute all these values. So i becomes integral root t dt by a and this is Again equal to 1 upon a integral of t to the power 1 by 2 dt Now this is equal to 1 upon a Integral of t to the power 1 by 2 is t to the power 1 by 2 plus 1 upon 1 by 2 plus 1 Plus c as the integral of x to the power n dx is x to the power n plus 1 upon n plus 1 Plus c. So here n is 1 by 2 So this is equal to 1 upon a Into t to the power 3 by 2 upon 3 by 2 plus c Which is again equal to 2 by 3 a into t to the power 3 by 2 plus c Let us now substitute the value of t here. So this becomes 2 by 3 a ax plus b To the power 3 by 2 plus c Hence the integral of the given function is 2 by 3 a ax plus b Whole to the power 3 by 2 plus c and this completes the question. Bye for now. Take care. Have a good day