EXAMPLE 4 Use the Table of Integrals to find SxVx² + 4x + 11dx. Since the table gives forms involving a² + x², V a² - x², and Vx - a?, but not v ax² + bx + c, we first complete the square. x? + 4x + 11 = (x + 2)² + 7 If we make the substitution u = x + 2 (so x = u - 2), the integrand will involve the pattern va + u². SxVx + 4x + 11dx = S( Vu– 2 x )Vư² + 7du = S u² + 7du – 2 | × Vư² + 7du - 2 SVūt + 7du The first integral is evaluated using the substitutiont = u² + 7. 1 Suvu + 7du =SVtdt 2 t (u² + 7) 3 For the second integral we use Formula 21 with a = v7. 7. Vu? + 7du = 5Vu² + 7+;In (u + Vu² +7) II

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
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Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 16E
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EXAMPLE 4
Use the Table of Integrals to find SxVx? + 4x + 11dx.
- x², va? - x², and
Since the table gives forms involving v a² +
Vx? - a2, but not v ax? + bx + c, we first complete the square.
SOLUTION
x2 + 4x + 11
= (x + 2)2 +
If we make the substitution u = x + 2 (so x = u - 2), the integrand will involve the
pattern va + u².
SxVx? + 4x + 11dx = S( Vu – 2
x )Vu² + 7du
= S Vu? + 7du – 2
x Vu2 + 7du - 2
Svū + 7du
+ 7.
The first integral is evaluated using the substitution t = u?
Suvu + 7du =-
_SVtdt
2
1
|(1² + 7) 3
For the second integral we use Formula 21 with a =
v7.
P+ iau=V교+7+2n (u+ v0P + 7)
V
Thus
x + 2
· 2
SxVx2 + 4x + 11dx =
(x2 + 4x + 11)
+ 4x + 11 -
In(x + 2 + Vx² + 4x + 11) + C
Transcribed Image Text:EXAMPLE 4 Use the Table of Integrals to find SxVx? + 4x + 11dx. - x², va? - x², and Since the table gives forms involving v a² + Vx? - a2, but not v ax? + bx + c, we first complete the square. SOLUTION x2 + 4x + 11 = (x + 2)2 + If we make the substitution u = x + 2 (so x = u - 2), the integrand will involve the pattern va + u². SxVx? + 4x + 11dx = S( Vu – 2 x )Vu² + 7du = S Vu? + 7du – 2 x Vu2 + 7du - 2 Svū + 7du + 7. The first integral is evaluated using the substitution t = u? Suvu + 7du =- _SVtdt 2 1 |(1² + 7) 3 For the second integral we use Formula 21 with a = v7. P+ iau=V교+7+2n (u+ v0P + 7) V Thus x + 2 · 2 SxVx2 + 4x + 11dx = (x2 + 4x + 11) + 4x + 11 - In(x + 2 + Vx² + 4x + 11) + C
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