Page 115 - Calculus Demystified

The Integral
CHAPTER 4
102
then we would see that we are integrating an expression of the form

ϕ (x)
ϕ(x)
(which we in fact encountered among our differentiation rules in Section 2.5).
As we know, expressions like this arise from differentiating log ϕ(x).
2
Returning to the original problem, we pose our initial guess as log[x + 3].
2
Differentiation of this expression gives the answer 2x/[x + 3]. This is close
to what we want, but we must adjust by a factor of 1/2. We write our final
answer as

x 1 2
dx = log[x + 3]+ C.
2
x + 3 2
You Try It: Calculate the indefinite integral
2
xe 3x +5 dx.

EXAMPLE 4.3
Calculate the indefinite integral

3 2 50 2
(x + x + 1) · (6x + 4x) dx.
SOLUTION
2 3
We observe that the expression 6x + 4x is nearly the derivative of x +
2
3
2
x + 1. In fact if we set ϕ(x) = x + x + 1 then the integrand (the quantity
that we are asked to integrate) is
50
[ϕ(x)] · 2ϕ (x).
51
It is natural to guess as our antiderivative [ϕ(x)] . Checking our work, we
find that
51 50
([ϕ(x)] ) = 51[ϕ(x)] · ϕ (x).
We see that the answer obtained is quite close to the answer we seek; it is off by
a numerical factor of 2/51. With this knowledge, we write our final answer as
2
3
2
3
2
2
(x + x + 1) 50 · (6x + 4x) dx = ·[x + x + 1] 51 + C.
51
You Try It: Calculate the indefinite integral
2
x
dx.
3
x + 5