OpenStax-CNX module: m47631
1
Techniques of integration:
∗
Completing the Square and Arctan
John Taylor
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
Abstract
completion square substitution integral arctan
1 Techniques of integration: Completing the Square and Arctan
(The small steps below can be used for a self test. To do so, Scroll in small increments.)
Find
I=
R
1
x2 +6x+10
∗ dx
.
Equation 1
How can we proceed?
By completing the square, we sometimes can match an integrand with those leading to the inverse trig
functions. Here we can work separately with the expression under the radical,
x2 + 6x + 10
Completing the square involves adding and subtracting the same quantity, followed by regrouping.
What value could we add or subtract inside the parentheses to make a perfect square?
If we subtract 1 to make 9, the quadratic would be a perfect square.
Do that.
x2 + 6x + 10 = x2 + 6x + 10 − 1 + 1
Notice that we also add 1 to compensate.
Simplify this.
x2 + 6x + 10 = x2 + 6x + 9 + 1
Factor the quadratic.
x2
+ 6x + 10 = x2 + 6x + 9 + 1
=
(x + 3)
2
+
1
Substitute this in Equation 1.
I=
R
1
x2 +6x+10
∗ dx =
R
1 2
(x+3)
+ 1 ∗ dx Equation 2
What should we do next?
We can try a change of variable by substituting
With this choice, express
du
du
in terms of
= dx
Use these results to rewrite Equation 2.
We get
I=
R
1
u2 +1
∗ du
What general integral can we use here?
∗
†
Version 1.1: Sep 11, 2013 12:43 pm +0000
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x.
u=x+3
OpenStax-CNX module: m47631
2
We may be able to use the integral leading to the arctan:
du
R
a2 +u2
= arctan
u
a
+C
What value should we use for
We should use
a
?
a=1
Do that.
I = arctan
= arctan
u
1
+C
(u) + C
Are we done?
No.
What else is required?
As always, we need to express the result in terms of the original variable.
Do that.
I = arctan (x + 3) + C
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