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Calculus Tutor here! It seems you got #3 right from the comments, so I'll focus on the other two. Problem 1: 1) For the first problem, foiling the (x - 1) with the (x^2 + x + 1) seems like the quickest way to simplify the problem before integration. 2) Doing this should result with x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1. 3) So now we have the integral from 1 to 3 of x^3 - 1 dx. 4) Using the power rule for each element should give us ( x^4 / 4 ) - x evaluated from 1 to 3. 5) Use the second fundamental theorem of calculus to get ( ((3)^4 / 4) - (3) ) - ( ((1)^4 / 4) - (1) ). I will leave you to figure out what that's equal to from there. Problem 2: 1) Do you know how to U substitute? This is a U substitution problem. 2) Let U = x^2 + 6x + 2 3) This means that du = 2x + 6 dx. 4) Using algebra we get that dx = du / (2 (x + 3)) 5) Making our substitution we end up with an integral of 1/2 (u)^(-1/2). Note that the x+3 from the dx substitution canceled out with the x+3 in the numerator of the fraction. 6) Use standard power rule to get u^(1/2). Substitute your x back in and you end up with (x^2 + 6x + 2)^(1/2), evaulated from 0 to 1. 7) With the second fundamental theorem of calculus, this becomes: ( ( (1)^2 + 6(1) + 2)^(1/2) ) - ( ( (0)^2 + 6(0) + 2)^(1/2) ) Again, I will leave this to you to evaluate fully. Let me know if that helps or if I can explain anything better :)

Do your own homework, kiddo.

Think I've gotten all the way through #3 correctly. Made u=5X+7, du=5dx, du/5=dx =Integral 17 to 37 (1/u)(du/5) =1/5 Integral 17 to 37 (1/u)(du) =1/5 ln [b]l[/b]u[b]l[/b] g(37) - g(17)= 1/5 ln [b]l[/b]37[b]l[/b] - 1/5 ln [b]l[/b]17[b]l[/b] =0.15554 Don't care if nobody reads this, Flood learn some math.