Let u = 2x + 3
x = (u-3)/2
du = 2dx
1. ∫x√(2x+3) = 1/4 integral[(u-3)sqrt(u)du]
= 1/4 integral[u^(3/2)- 3u^(1/2)]du
2. See attachment
Working on a Calculus I review and two integration problems have stumped me:
Help would be much appreciated because I don't know where to begin on these and have a strong belief that they will be on my quiz tomorrow.
for the second integral notice that the derivative of the denominator is 4x+2. if the numerator is 4x+2, it would just become a simple u substitution of form 1/u du. so multiply the numerator by 2 (but you also have to divide by 2 and you can factor that out of the integral). the numerator becomes 4x-10. what you want is 4x+2 so add 12 to 4x-10 to get 4x-10+12 = 4x+2. but you also have to subtract 12 so you would have (4x+2)-12. now split the integral up as (4x+2)/(2x²+2x+2) - 12/(2x²+2x+2). use u substitution on the first integral and complete the square on the denominator of the 2nd integral.