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Unformatted text preview: l problem is equal to
tan2 θ sec θdθ.
We use integration by parts. Let u = tan θ and dv = sec θ tan θdθ . It follows that
du = sec2 θdθ and v = sec θ . So
tan2 θ sec θdθ = sec θ tan θ − sec3 θdθ = sec θ tan θ − = sec θ tan θ −
Add sec θdθ − (tan2 θ + 1) sec θdθ tan2 θ sec θdθ. tan2 θ sec θdθ to both sides to see that
2 tan2 θ sec θdθ = sec θ tan θ − sec θdθ. So
x2 + 2xdx = tan2 θ sec θdθ = (1/2) sec θ tan θ − sec θdθ = (1/2) [sec θ tan θ − ln  sec θ + tan θ ] + C
= (1/2) (x + 1) x2 + 2x − ln (x + 1) + x2 + 2 x + C . Check. The derivative of
(1/2) (x + 1) x2 + 2x − ln[(x + 1) +
is
(x + 1)(2x + 2)
√
(1/2)
+
2 x2 + 2 x
(x + 1)2
+
= (1/2) √ x2 x2 + 2x − + 2x − x2 + 2 x 1+
(x + 1)
1+ 2
√ x+2
2 x2 +2x
√
+ x2 + √ x+1
x2 +2x √ 2x 2 (x + 1)2
+
= (1/2) √
x2 + 2 x x2 + 2 x − (x + 1)2
+
= (1/2) √
x2 + 2 x √ x2 + 2 x + x + 1
√
√
[(x + 1) + x2 + 2x] x2 + 2x
x2 + 2 x − √ 1
x2 + 2 x 1
=√
(x + 1)2 + x2 + 2x − 1
2 + 2x
2x
1
=√
2 x2 + 4 x
2 + 2x
2x
=√ 1
x2 + 2 x =
2 + 2x
x x2 + 2x....
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This document was uploaded on 03/23/2014 for the course MATH 142 at South Carolina.
 Fall '11
 KUSTIN
 Calculus

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