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x1J:(x2+2x)2 dx=(x2+2x)3C(x2 2x)3C+3(x2+2x)3+C622-1C+-2 2x)1+C

Question
x1
J:
(x2+2x)2 dx=
(x2+2x)3
C
(x2 2x)3
C
+
3
(x2+2x)3
+C
6
22-1C
+
-2 2x)1+C
help_outline

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x1 J: (x2+2x)2 dx= (x2+2x)3 C (x2 2x)3 C + 3 (x2+2x)3 +C 6 22-1C + -2 2x)1+C

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check_circleAnswer
Step 1
The given integral is fx+1
dx
'(x+2x)
Use the substitutions to solve the integral
then dx = du andx =u -1
Let u x
help_outline

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The given integral is fx+1 dx '(x+2x) Use the substitutions to solve the integral then dx = du andx =u -1 Let u x

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Step 2
Simplify the term (x +2x)
as follows
(x21) -(u-1)+2(u-1)
- (u-2+1+2-2)
(u-1)
((a-b)=a* -2ab+b°)
Thus, the given integral becomes
x+1
(:u =x+1 and (x +2x) = (u* -1))
du
(uP-1)
'(x+2)
help_outline

Image Transcriptionclose

Simplify the term (x +2x) as follows (x21) -(u-1)+2(u-1) - (u-2+1+2-2) (u-1) ((a-b)=a* -2ab+b°) Thus, the given integral becomes x+1 (:u =x+1 and (x +2x) = (u* -1)) du (uP-1) '(x+2)

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Step 3
Let v u-1 then,
dv 2u
dv
2
x1
du as follows
Integrate
2x
dv
dv
2-1 and u = °
2
r=f
(x+2x)
(v)
dv
2v
1r1
dv
dv
1v24
+C
2-2 1
n 1
help_outline

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Let v u-1 then, dv 2u dv 2 x1 du as follows Integrate 2x dv dv 2-1 and u = ° 2 r=f (x+2x) (v) dv 2v 1r1 dv dv 1v24 +C 2-2 1 n 1

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Calculus

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