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How do you integrate #int (x^3+2)dx#?

Answer 1

Samantha Adkison

#intx^3+2dx#

#=(x^(3+1))/(3+1)+2x+C#

#=1/4x^4+2x+C#

This is due to:

#intx^ndx#

#=(x^(n+1))/(n+1)+C#

And also:

#intkdx#

#=kx+C#

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Answer 2

Evelyn Agard

#int(x^3+2)dx=x^4/4+2x+c#

Using the power rule of integration, #intx^ndx=x^(n+1)/(n+1)+c,#

#int(x^3+2)dx=int(x^3+2x^0)#

#=x^(3+1)/(3+1)+(2x^1)/1+c#

#=x^4/4+2x+c#

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Answer 3

Ezra Adams

To integrate ( \int (x^3+2) , dx ), you would follow these steps:

Find the antiderivative of each term: [ \int x^3 , dx = \frac{x^4}{4} + C ] [ \int 2 , dx = 2x + C ]
Combine the antiderivatives: [ \int (x^3+2) , dx = \frac{x^4}{4} + 2x + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.