Expert: Alon Mandes - 5/17/2009

Question
Compute each integral: 1) ∫b=4 a=1 √(1+√(x))dx   and

                      2) ∫(3x+1)/(2x+5)dx
Number one is definite integral, and two is indefinite.
Can ypu please explain.  Thanks.

Answer
1. ∫√[1+√x] dx
In this section we will use the substitute method :
We set t=1+√x that fives us dt=dx/(2√x) which means dx=2(t-1).Then
our integral becomes :
∫√[1+√x] dx = 2∫(t-1)√t dt = 2∫t^(3/2) dt - 2∫√t dt =
= 2*(2/5)t^(5/2) - 2*(2/3)t^(3/2) . Thus,
∫√[1+√x] dx = (4/5)[1+√x]^(5/2)-(4/3)[1+√x]^(3/2) =
4[1+√x]^(3/2) * { (1/5)[1+√x]-(1/3)] =
4[1+√x]^(3/2) * { (1/5)+(1/5)√x-(1/3)] =
(4/15)[1+√x]^(3/2) * (3√x-2) =
(4/15)(3√x-2)[1+√x]^(3/2)

Now we calculate the area using the limits a&b:

b
∫√[1+√x] dx = (4/15){(3√b-2)[1+√b]^(3/2)-(3√a-2)[1+√a]^(3/2)}
a

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2. ∫(3x+1)/(2x+5) dx
In this section we will make mathematical "cosmetics" to the
fraction in order to get form that is easily integrated :
(3x+1)/(2x+5) = (2x+5+x-4)/(2x+5) = 1 + (x-4)/(2x+5) =
1 + x/(2x+5) - 4/(2x+5) = 1 + (½)x/(x + 5/2) - 4/(2x+5) =
1 + (½)(x+ 5/2)/(x + 5/2) - (½)(5/2)/(x + 5/2) - 4/(2x+5) =
1 + (½) - (½)(5/2)/(x + 5/2) - 4/(2x+5)  .
We know that ∫a/(b+cx) dx = (a/c)Ln(b+cx) . Therefore :
∫(3x+1)/(2x+5) dx = x+½x-(5/4)Ln[x + 5/2]-4Ln[2x+5] =
x+½x-(5/4)Ln[(2x+5)/2]-4Ln[2x+5] =
x+½x-(5/4)Ln[2x+5]-(5/4)Ln[2]-4Ln[2x+5] .

Alon.