This video from IntegralCALC shows you how to solve the 6th Integrals Math problem.
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Integrals Example Number 6 The next thing that we’re going to do is the first one involving a square root so we’re going to have x5/2- 5/x4vx dx. So the first thing we’re going to do is try to simplify this x5 this is going to stay the same we’re going to eliminate this fractions by moving the x4 of the top what do we do that as a change the sign on the x sign from a positive to a negative, 5x to -4 same as 5/x4 and then of course we know rule the vx is the same as x1/2. So we’re going to say –x1/2dx, make this easier. So now we can take the integral x 5/2 we of course add 1 so we’re going to say + 2/2 on the exponent is of course 7/2 and then we need to divide one which is the coefficient by the new exponent 7/2 then –x add one to the exponent -4+1= -3 and then divide the coefficient by the new exponent 5/-3 and then x ½ + 1 2/2 is of course 3/2, 1+2 =3 over 2 and then 1 is the coefficient that’s implied here divided by the new exponent which 3/2, 1/3/2 and then +b. So now we have our integral and all we have to do is simplify it. So 1/7/2 as mentioned in the last video instead of dividing by fraction we can multiply it by inverse so instead of 1/7/2 that’s the same thing as 1times the inverse of the fraction 2/7, 1*2/7 is 2/7 so 2/7x7/2 we have a negative here on the bottom so a negative times a negative 5/3 is going to be a positive so +5/3 and then x-3 we never want to have negative exponent and the answer so we are going to move the x to the bottom and change the sign on the exponent. So instead of x-3 and the numerator we’re going to have x+3 in the denominator. So 5/3x3 and then 1/3/2 instead of 1 ÷ 3/2 we’re going to do 1* 2/3 which of course is 2/3 – 2/3x3/2 + c and that’s the answer.