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ʃ (2x + 6)(x2 + 6x + 3)7 dx

Guest Mar 16, 2017
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 #1
avatar+18619 
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ʃ (2x + 6)(x2 + 6x + 3)7 dx

 

\(\begin{array}{|rcll|} \hline && \int (2x + 6)(x^2 + 6x + 3)\cdot 7\ dx \\ &=& 7\cdot \int (2x + 6)(x^2 + 6x + 3)\ dx \\ &=& 7\cdot \int (2x^3+12x^2+6x+6x^2+36x+18)\ dx \\ &=& 7\cdot \int (2x^3+18x^2+42x+18)\ dx \\ &=& 7\cdot \left( \frac24x^4+\frac{18}{3}x^3+\frac{42}{2}x^2 + 18x \right) \\ &=& 3.5x^4+42x^3+147x^2+126 + c \\ \hline \end{array} \)

 

laugh

heureka  Mar 16, 2017
 #3
avatar+26242 
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heureka's 126 should be 126x of course!

Alan  Mar 16, 2017
 #4
avatar+18619 
0

ʃ (2x + 6)(x2 + 6x + 3)7 dx

 

Sorry, without mistakes:

 

\(\begin{array}{rcll} \hline && \int (2x + 6)(x^2 + 6x + 3)\cdot 7\ dx \\ \hline \end{array}\)

 

\(\begin{array}{rcll} &=& 7\cdot \int (2x + 6)(x^2 + 6x + 3)\ dx \\ &=& 7\cdot \int (2x^3+12x^2+6x+6x^2+36x+18)\ dx \\ &=& 7\cdot \int (2x^3+18x^2+42x+18)\ dx \\ &=& 7\cdot \left( \frac24x^4+\frac{18}{3}x^3+\frac{42}{2}x^2 + 18x \right) \\ &=& 3.5x^4+42x^3+147x^2+126x + c \\ \end{array}\)

 

 

blushlaugh

heureka  Mar 16, 2017
 #2
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Take the integral:
 integral7 (2 x + 6) (x^2 + 6 x + 3) dx
Factor out constants:
 = 7 integral(2 x + 6) (x^2 + 6 x + 3) dx
Expanding the integrand (2 x + 6) (x^2 + 6 x + 3) gives 2 x^3 + 18 x^2 + 42 x + 18:
 = 7 integral(2 x^3 + 18 x^2 + 42 x + 18) dx
Integrate the sum term by term and factor out constants:
 = 14 integral x^3 dx + 126 integral x^2 dx + 294 integral x dx + 126 integral1 dx
The integral of x^3 is x^4/4:
 = (7 x^4)/2 + 126 integral x^2 dx + 294 integral x dx + 126 integral1 dx
The integral of x^2 is x^3/3:
 = 42 x^3 + (7 x^4)/2 + 294 integral x dx + 126 integral1 dx
The integral of x is x^2/2:
 = 147 x^2 + 42 x^3 + (7 x^4)/2 + 126 integral1 dx
The integral of 1 is x:
 = (7 x^4)/2 + 42 x^3 + 147 x^2 + 126 x + constant
Which is equal to:
Answer: |= 14 (x^4/4 + 3 x^3 + (21 x^2)/2 + 9 x) + constant

Guest Mar 16, 2017

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