Calculus II
Contents
Series
Integrals
Definite integrals
Some primitives
Numerical methods
Exercices
© The scientific sentence. 2010
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Calculus II: Rational functions primitives
1. Primitive involving the completion square method
1.1. ∫ dx/(ax2 + bx + c)
Let the trinomial:
ax2 + bx + c
That we can write:
a[x2 + 2 (b/2a)x + (b/2a)2 - (b/2a)2 + (c/a)]
=
a[(x + b/2a)2 + (c/a) - (b/2a)2]
Let change the variable x into
u = x + b/2a , so du = dx
And let
(c/a) - (b/2a)2 = k2
we have then
If (c/a) - (b/2a)2 > 0
ax2 + bx + c = a[u2 + k 2]
If (c/a) - (b/2a)2 < 0
ax2 + bx + c = a[u2 - k 2]
Therefore:
If (c/a) - (b/2a)2 > 0
ax2 + bx + c = a[u2 + k 2]
∫ dx/(ax2 + bx + c) = ∫ du/a[u2 + k 2] = (1/ka) arctan(u/k) =
[1/a((c/a) - (b/2a)2)1/2] arctan[(x + b/2a)/((c/a) - (b/2a)2)1/2]
If (c/a) - (b/2a)2 < 0
ax2 + bx + c = a[u2 - k 2]
∫ dx/(ax2 + bx + c) = ∫ du/a (u - k)(u + k) =
(1/a) [∫ cst1 du/(u - k) + ∫ cst2 du/(u + k)]
(by using the partial fraction decomposition).
P(x) = ax2 + bx + c
(c/a) - (b/2a)2 = ± k2
u = x + b/2a
du = dx
If (c/a) - (b/2a)2 > 0
ax2 + bx + c = a[u2 + k 2]
∫ dx/(ax2 + bx + c) =
∫ du/a[u2 + k 2] = (1/ka) arctan(u/k) + cst.
If (c/a) - (b/2a)2 < 0
ax2 + bx + c = a[u2 - k 2]
∫ dx/(ax2 + bx + c) = ∫ du/a (u - k)(u + k) =
(1/a) [∫ cst1 du/(u - k) + ∫ cst2 du/(u + k)]
1.2. Example 1: (c/a) - (b/2a)2 > 0
P(x) = x2 + 4x + 13
a = 1 , b = 4, c = 13
k2 = (c/a) - (b/2a)2 =
13 - 4 = 9 > 0 .
k2 = + 9
u = x + b/2a = x + 2
Therefore
∫ dx/(x2 + 4x + 13) = (1/ka) arctan(u/k) + cst =
(1/3) arctan[(x + 2)/3] + cst.
∫ dx/(x2 + 4x + 13) = (1/3) arctan[(x + 2)/3] + cst.
1.3. Example 2: (c/a) - (b/2a)2 < 0
∫ 5 dx /(x2 - 2 x - 3)
For the polynomial: x2 - 2 x - 3
(c/a) - (b/2a)2 = - 3 - (- 2/2)2 = - 4 < 0
x2 - 2 x - 3 = x2 - 2 x + 1 - 1 - 3 =
(x - 1)2 - 4 = (x - 1 - 2)(x - 1 + 2) = (x - 3)(x + 1)
And the partial functions decomposition method:
1/(x - 3)(x + 1) = m/(x - 3) + n/(x + 1) =
[(m + n) x + (m - 3n)]/(x - 3)(x + 1)
Equating the numerators, we get:
(m + n) x + (m - 3n) = 1. That is
m + n = 0, and
m - 3n = 1
That give:
m = + 1/4
n = - 1/4
Therefore
1/(x - 3)(x + 1) = (1/4)[1/(x - 3) - 1/(x + 1)]
Hence
∫ (5 dx/(x2 - 2 x - 3)
= 5 ∫ (dx/(x2 - 2 x - 3) =
(5/4) ∫ dx (1/(x - 3) - 1/(x + 1)) =
(5/4) [(ln|x - 3|) - ln|x + 1|] =
(5/4) (ln |x - 3|/|x + 1|) + cst.
We have then:
∫ (5 dx/(x2 - 2 x - 3) = (5/4) (ln |x - 3|/|x + 1|) + cst
1.4. ∫ (m x + n)dx/(ax2 + bx + c)
∫ (x + 4)dx/(x2 - 2 x - 3 )
x + 4 = x - 1 + 5
(x + 4)/(x2 - 2 x - 3) =
(x - 1)/(x2 - 2 x - 3) +
5/(x2 - 2 x - 3)
With u = x2 - 2 x - 3, so du = 2 x - 2 = 2(x - 1), we
have:
(x - 1)/(x2 - 2 x - 3) = (1/2) du/u
so
∫ (x - 1)dx/(x2 - 2 x - 3) = (1/2) ∫du/u =
(1/2) ln|u| + cst = (1/2) ln|x2 - 2 x - 3| + cst
∫ (x - 1)dx/(x2 - 2 x - 3) =
(1/2) ln|x2 - 2 x - 3| + cst
From the example 1.3 above, we have found:
∫ (5 dx/(x2 - 2 x - 3) = (5/4) (ln |x - 3|/|x + 1|) + cst
Finally,
∫ (x + 4)dx/(x2 - 2 x - 3 ) =
∫ (x - 1)dx/(x2 - 2 x - 3) + ∫ (5 dx/(x2 - 2 x - 3) =
(1/2) ln|x2 - 2 x - 3| + (5/4) (ln |x - 3|/|x + 1|) + cst
4. Exercise
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