MATH 106A,B - CALCULUS II WINTER 2008
QUIZ 5
NAME:
Show ALL your work CAREFULLY.
Use the method of completing the square together with the technique of trigonometric substitution to find
the following indefinite integral
Zdx
√x2+ 2x+ 26
.
(useful fact: Rsec x dx = ln |sec x+ tan x|+C)
By completing the square, we write x2+ 2x+ 26 = (x2+ 2x+ 1) + 25 = (x+ 1)2+ 52. Now we let
u=x+ 1 so that du =dx and
Zdx
√x2+ 2x+ 26 =Zdu
√u2+ 52
.
We use the trigonometric substitution u= 5 tan θ. Thus, du = 5 sec2θ dθ and √u+52=p52(tan2θ+ 1) =
5 sec θ. It follows that
Zdu
√u2+ 52=Z5 sec2θ dθ
5 sec θ=Zsec θ dθ
= ln |sec θ+ tan θ|+C
= ln |√u2+ 52
5+u
5|+C
= ln |pu2+ 52+u|+C.
Hence,
Zdx
√x2+ 2x+ 26 = ln |px2+ 2x+ 26 + x+ 1|+C.
Date: February 13, 2008.
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